Oscillations and Waves MCQs with Answers – Part 7 (NEET Physics)

Explanation: \(\textbf{Competing spring and apparatus forces:}\) The spring exerts a force \(-kx\), while the additional apparatus exerts \(+\alpha x\). Their resultant is \[ F_{\mathrm{net}}=-(k-\alpha)x \] The sign of the effective coefficient is crucial because a restoring force must act towards equilibrium. \(\textbf{Stable oscillation with reduced stiffness:}\) When \(\alpha=36\,\text{N m}^{-1}\), the effective stiffness is \(100-36=64\,\text{N m}^{-1}\). With mass \(0.25\,\text{kg}\), the angular frequency is \[ \omega=\sqrt{\frac{64}{0.25}}=16\,\text{rad s}^{-1} \] so Statement I is correct. When \(\alpha=100\,\text{N m}^{-1}\), the apparatus exactly cancels the spring force at every displacement. \(\textbf{Zero-net-force case:}\) A block released from rest then remains where it is placed, rather than oscillating about a restoring equilibrium. Statement II is correct. With \(\alpha=144\,\text{N m}^{-1}\), the net force becomes \(+44x\). \(\textbf{Unstable equilibrium when outward force dominates:}\) This force acts away from \(x=0\) on both sides, making that position an unstable equilibrium. It does not produce SHM, so Statement III is false. For \(0\lt \alpha\lt 100\,\text{N m}^{-1}\), the angular frequency is \[ \omega=\sqrt{\frac{100-\alpha}{0.25}} \] \(\textbf{How the frequency depends on apparatus strength:}\) Increasing \(\alpha\) decreases the remaining restoring coefficient and therefore lowers angular frequency, making Statement IV correct. A position-dependent force need not be restoring merely because its magnitude grows with displacement. Its direction relative to equilibrium determines its effect. The apparatus opposes the spring's restoring action in this case, so Statement V is false.

Explanation: \(\textbf{Interpreting the slope of the velocity-displacement curve:}\) The plotted variables are instantaneous velocity and displacement. By the chain rule, \[ \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt} \] Since \(dx/dt=v\), \[ a=v\frac{dv}{dx} \] Thus, the tangent slope of the velocity-displacement curve gives acceleration only after multiplication by velocity. \(\textbf{Acceleration at the marked point:}\) The measured values are \[ x=+0.060\,\text{m} \] \[ v=-0.80\,\text{m s}^{-1} \] \[ \frac{dv}{dx}=+7.5\,\text{s}^{-1} \] Therefore, \[ a=(-0.80)(+7.5) \] \[ a=-6.0\,\text{m s}^{-2} \] The negative acceleration is directed towards equilibrium, as expected for positive displacement. \(\textbf{Recovering angular frequency:}\) For SHM, \[ a=-\omega^2x \] Therefore, \[ \omega^2=-\frac ax \] Substituting, \[ \omega^2=-\frac{-6.0}{0.060} \] \[ \omega^2=100\,\text{s}^{-2} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Determining amplitude:}\) Use the SHM speed relation: \[ v^2=\omega^2(A^2-x^2) \] Thus, \[ A^2=x^2+\frac{v^2}{\omega^2} \] \[ A^2=(0.060)^2+\frac{(0.80)^2}{(10)^2} \] \[ A^2=0.0036+0.0064 \] \[ A^2=0.0100\,\text{m}^2 \] \[ A=0.10\,\text{m} \] \(\textbf{Spring force constant:}\) For a spring-mass oscillator, \[ \omega^2=\frac km \] Therefore, \[ k=m\omega^2 \] \[ k=0.50(100) \] \[ k=50\,\text{N m}^{-1} \] \(\textbf{Energy fraction represented by motion:}\) The kinetic fraction is \[ \frac KE=\frac{v^2}{\omega^2A^2} \] Substituting, \[ \frac KE=\frac{0.64}{100(0.10)^2} \] \[ \frac KE=0.64 \] \(\textbf{Physical interpretation:}\) A velocity-displacement graph carries different information from a displacement-time graph. Its tangent slope is \(dv/dx\), not instantaneous velocity or acceleration by itself. Multiplying that slope by the signed velocity gives the instantaneous acceleration. The SHM restoring relation then determines angular frequency, while the marked position and speed determine amplitude. The spring force constant and energy fraction follow from these recovered physical quantities. The positive graph slope on the negative-velocity branch is fully consistent with a negative restoring acceleration. \(\textbf{Final answer:}\) \[ \boxed{\omega=10\,\text{rad s}^{-1},\qquad A=0.10\,\text{m},\qquad k=50\,\text{N m}^{-1},\qquad\frac KE=0.64} \]

Explanation: \(\textbf{Reconstructing the displacement from the initial state:}\) The general SHM displacement may be written as \[ x(t)=C\cos(\omega t)+D\sin(\omega t) \] At \(t=0\), \[ x(0)=C \] Therefore, \[ C=+0.060\,\text{m} \] Differentiating, \[ v(t)=-C\omega\sin(\omega t)+D\omega\cos(\omega t) \] Initially, \[ v(0)=D\omega \] Thus, \[ D=\frac{v(0)}{\omega} \] \[ D=\frac{0.80}{10} \] \[ D=0.080\,\text{m} \] The complete displacement relation is \[ x(t)=0.060\cos(10t)+0.080\sin(10t) \] \(\textbf{Motion after one-quarter period:}\) A quarter period corresponds to the phase change \[ \omega\frac T4=\frac{\pi}{2} \] Therefore, \[ x(T/4)=0.060\cos\frac{\pi}{2}+0.080\sin\frac{\pi}{2} \] \[ x(T/4)=+0.080\,\text{m} \] The velocity expression is \[ v(t)=-0.60\sin(10t)+0.80\cos(10t) \] At \(T/4\), \[ v(T/4)=-0.60\,\text{m s}^{-1} \] \(\textbf{Acceleration at the quarter-period observation:}\) For SHM, \[ a=-\omega^2x \] Consequently, \[ a(T/4)=-(10)^2(0.080) \] \[ a(T/4)=-8.0\,\text{m s}^{-2} \] \(\textbf{Motion after half a period:}\) Half a period corresponds to phase advance \(\pi\). Therefore, \[ x(T/2)=0.060\cos\pi+0.080\sin\pi \] \[ x(T/2)=-0.060\,\text{m} \] Similarly, \[ v(T/2)=-0.60\sin\pi+0.80\cos\pi \] \[ v(T/2)=-0.80\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) The initial displacement and velocity together determine the complete harmonic state. A quarter-period advance exchanges the roles of the initial displacement and velocity after the appropriate angular-frequency scaling and sign change. After half a period, both signed displacement and signed velocity reverse. The acceleration also reverses because it is proportional to negative displacement. These relations hold for any SHM phase and do not require the particle to begin at a turning point. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} t=T/4:&\quad x=+0.080\,\text{m},\quad v=-0.60\,\text{m s}^{-1},\quad a=-8.0\,\text{m s}^{-2}\\ t=T/2:&\quad x=-0.060\,\text{m},\quad v=-0.80\,\text{m s}^{-1} \end{aligned}} \]

Explanation: \(\textbf{Initial oscillation frequency and equilibrium speed:}\) The initial total oscillating mass is \[ M_i=0.50+0.50 \] \[ M_i=1.00\,\text{kg} \] Its angular frequency is \[ \omega_i=\sqrt{\frac{k}{M_i}} \] \[ \omega_i=\sqrt{\frac{200}{1.00}} \] \[ \omega_i=10\sqrt2\,\text{rad s}^{-1} \] At equilibrium, the oscillator's speed is maximum: \[ v_i=\omega_iA_i \] \[ v_i=10\sqrt2(0.10) \] \[ v_i=\sqrt2\,\text{m s}^{-1} \] \(\textbf{State immediately after detachment:}\) The ideal mechanism imparts no impulse. The remaining spring-attached block therefore retains velocity \[ v_0=+\sqrt2\,\text{m s}^{-1} \] The separation occurs at equilibrium, so the spring extension is zero. The remaining block begins its new SHM at equilibrium with this unchanged speed. \(\textbf{Frequency and period after the reduction in mass:}\) The mass remaining on the spring is \[ M_f=0.50\,\text{kg} \] Therefore, \[ \omega_f=\sqrt{\frac{k}{M_f}} \] \[ \omega_f=\sqrt{\frac{200}{0.50}} \] \[ \omega_f=20\,\text{rad s}^{-1} \] The ratio of periods is \[ \frac{T_f}{T_i}=\frac{\omega_i}{\omega_f} \] \[ \frac{T_f}{T_i}=\frac{10\sqrt2}{20} \] \[ \frac{T_f}{T_i}=\frac1{\sqrt2} \] \(\textbf{New amplitude of the remaining oscillator:}\) At an equilibrium crossing, \[ v_{\max}=\omega_fA_f \] Thus, \[ A_f=\frac{v_0}{\omega_f} \] \[ A_f=\frac{\sqrt2}{20} \] \[ A_f=\frac1{10\sqrt2}\,\text{m} \] \(\textbf{Energy carried by the detached block:}\) The detached block has mass \(0.50\,\text{kg}\) and speed \(\sqrt2\,\text{m s}^{-1}\). Its kinetic energy is \[ K_{\mathrm{detached}}=\frac12mv_0^2 \] \[ K_{\mathrm{detached}}=\frac12(0.50)(\sqrt2)^2 \] \[ K_{\mathrm{detached}}=0.50\,\text{J} \] \(\textbf{Checking the energy remaining in spring oscillation:}\) The original oscillation energy is \[ E_i=\frac12kA_i^2 \] \[ E_i=\frac12(200)(0.10)^2 \] \[ E_i=1.00\,\text{J} \] The final spring oscillator has energy \[ E_f=\frac12kA_f^2 \] \[ E_f=\frac12(200)\left(\frac1{10\sqrt2}\right)^2 \] \[ E_f=0.50\,\text{J} \] Thus, \[ E_i=E_f+K_{\mathrm{detached}} \] The energy balance is consistent with ideal impulse-free separation. \(\textbf{Physical interpretation:}\) Detaching a block at equilibrium does not require stopping either block. Both retain their instantaneous velocity at separation. The remaining block has less inertia and therefore oscillates with a higher angular frequency. Since its inherited speed is unchanged, its new amplitude becomes smaller. The apparent reduction in spring-oscillation energy is accounted for by kinetic energy carried away by the detached mass, not by dissipation within the ideal spring. \(\textbf{Final answer:}\) \[ \boxed{A_f=\frac1{10\sqrt2}\,\text{m},\qquad\frac{T_f}{T_i}=\frac1{\sqrt2},\qquad K_{\mathrm{detached}}=0.50\,\text{J}} \]

Explanation: \(\textbf{Recovering the force-displacement slope:}\) The restoring-force law has the form \[ F=-k(x-x_{\mathrm{eq}}) \] This can be written as \[ F=-kx+kx_{\mathrm{eq}} \] The slope of the force-coordinate relation is therefore \(-k\). From the two measurements, \[ \frac{\Delta F}{\Delta x} =\frac{-2.0-(+6.0)}{(+0.10)-(-0.10)} \] \[ \frac{\Delta F}{\Delta x}=-40\,\text{N m}^{-1} \] Hence, \[ k=40\,\text{N m}^{-1} \] \(\textbf{Determining the equilibrium coordinate:}\) Use the measurement at \(x=+0.10\,\text{m}\): \[ -2.0=-40(0.10-x_{\mathrm{eq}}) \] Expanding, \[ -2.0=-4.0+40x_{\mathrm{eq}} \] Therefore, \[ x_{\mathrm{eq}}=+0.050\,\text{m} \] The equilibrium position is displaced towards positive \(x\) from the laboratory origin. \(\textbf{Amplitude fixed by the release position:}\) The mass is released from rest at \[ x_0=-0.050\,\text{m} \] Its initial displacement from equilibrium is \[ z_0=x_0-x_{\mathrm{eq}} \] \[ z_0=-0.050-0.050 \] \[ z_0=-0.10\,\text{m} \] Since the initial velocity is zero, it starts at a negative turning point relative to equilibrium. Thus, \[ A=0.10\,\text{m} \] \(\textbf{Oscillation frequency from the measured force constant:}\) The angular frequency is \[ \omega=\sqrt{\frac km} \] \[ \omega=\sqrt{\frac{40}{0.40}} \] \[ \omega=10\,\text{rad s}^{-1} \] Therefore, \[ T=\frac{2\pi}{10} \] \[ T=\frac{\pi}{5}\,\text{s} \] \(\textbf{Time to the first equilibrium crossing:}\) A particle released from a turning point reaches equilibrium for the first time after one-quarter period. Hence, \[ t_{\mathrm{first}}=\frac T4 \] \[ t_{\mathrm{first}}=\frac{\pi}{20}\,\text{s} \] It crosses equilibrium moving towards positive \(x\). \(\textbf{Physical interpretation:}\) A linear force-coordinate graph can describe SHM even when the graph does not pass through the laboratory coordinate origin. The negative slope measures the restoring-force constant, whereas the zero-force intercept determines the true equilibrium position. The initial release point must be compared with that equilibrium rather than with \(x=0\). Using the corrected displacement identifies the amplitude and turning-point state. The mass and recovered stiffness then determine the period and the first equilibrium-crossing time. \(\textbf{Final answer:}\) \[ \boxed{k=40\,\text{N m}^{-1},\qquad x_{\mathrm{eq}}=+0.050\,\text{m},\qquad A=0.10\,\text{m},\qquad t_{\mathrm{first}}=\frac{\pi}{20}\,\text{s}} \]

Explanation: \(\textbf{Effective force in the accelerating platform frame:}\) Choose positive displacement relative to the platform towards positive \(x\). The spring force on the block is \[ F_s=-kx \] Because the platform accelerates towards positive \(x\), the platform frame includes a pseudo-force towards negative \(x\): \[ F_{\mathrm{pseudo}}=-ma_p \] The equation of relative motion is therefore \[ m\ddot x=-kx-ma_p \] \(\textbf{New equilibrium relative to the platform:}\) At relative equilibrium, \[ -kx_{\mathrm{eq}}-ma_p=0 \] Thus, \[ x_{\mathrm{eq}}=-\frac{ma_p}{k} \] Substituting, \[ x_{\mathrm{eq}}=-\frac{1.0(4.0)}{100} \] \[ x_{\mathrm{eq}}=-0.040\,\text{m} \] The equilibrium lies to the left of the spring's natural-length position. \(\textbf{Oscillation frequency and period:}\) Define displacement from the shifted equilibrium: \[ z=x-x_{\mathrm{eq}} \] The relative equation of motion becomes \[ m\ddot z=-kz \] Hence, \[ \omega=\sqrt{\frac km} \] \[ \omega=\sqrt{\frac{100}{1.0}} \] \[ \omega=10\,\text{rad s}^{-1} \] The period is \[ T=\frac{2\pi}{10} \] \[ T=\frac{\pi}{5}\,\text{s} \] \(\textbf{Amplitude from the switching conditions:}\) At the onset of platform acceleration, the spring remains at its natural length: \[ x(0)=0 \] The block also has zero initial velocity relative to the platform: \[ \dot x(0)=0 \] Relative to the new equilibrium, \[ z(0)=0-(-0.040) \] \[ z(0)=+0.040\,\text{m} \] Since relative velocity is zero, the block begins at a positive turning point of its new relative SHM. Therefore, \[ A=0.040\,\text{m} \] \(\textbf{Initial acceleration in the ground frame:}\) At the switching instant, the spring is unstretched. Its actual horizontal force on the block is therefore zero. In the inertial ground frame, \[ ma_{\mathrm{ground}}=F_s=0 \] Consequently, \[ a_{\mathrm{ground}}(0)=0 \] This differs from the platform's imposed acceleration. The block's initial acceleration relative to the platform is \[ a_{\mathrm{relative}}(0)=0-4.0 \] \[ a_{\mathrm{relative}}(0)=-4.0\,\text{m s}^{-2} \] \(\textbf{Maximum speed relative to the platform:}\) The relative motion is SHM of amplitude \(0.040\,\text{m}\) and angular frequency \(10\,\text{rad s}^{-1}\). Thus, \[ v_{\mathrm{rel,max}}=\omega A \] \[ v_{\mathrm{rel,max}}=10(0.040) \] \[ v_{\mathrm{rel,max}}=0.40\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) A uniformly accelerating platform creates a shifted equilibrium when motion is described in its non-inertial frame. The pseudo-force affects this equilibrium position but does not change the spring's restoring-force coefficient. Consequently, the relative oscillation frequency remains \(\sqrt{k/m}\). At the onset, the block has no actual horizontal force because the spring is still at its natural length. Its ground-frame acceleration is initially zero, even though its acceleration relative to the accelerating platform is negative. Distinguishing inertial and platform-relative accelerations is essential to the physical interpretation. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} x_{\mathrm{eq}}=-0.040\,\text{m},\quad A=0.040\,\text{m},\quad T=\frac{\pi}{5}\,\text{s}\\ a_{\mathrm{ground}}(0)=0,\quad v_{\mathrm{rel,max}}=0.40\,\text{m s}^{-1} \end{gathered}} \]

Explanation: \(\textbf{Recovering stiffness from the potential-energy graph:}\) The potential-energy minimum is \(0.40\,\text{J}\), rather than zero. For ideal SHM, an additive potential-energy constant does not affect the restoring force. Thus, \[ U(x)=0.40+\frac12kx^2 \] At \(x=+0.10\,\text{m}\), \[ 0.65=0.40+\frac12k(0.10)^2 \] \[ 0.25=0.005k \] \[ k=50\,\text{N m}^{-1} \] \(\textbf{Restoring-force direction at the negative displacement:}\) The restoring force is related to the potential-energy slope by \[ F=-\frac{dU}{dx}=-kx \] At \(x=-0.12\,\text{m}\), \[ F=-50(-0.12) \] \[ F=+6.0\,\text{N} \] The force is directed towards positive \(x\), namely towards equilibrium. \(\textbf{Recovering the kinetic energy from the turning point:}\) At a turning point, speed is zero, so total energy equals potential energy. With amplitude \(A=0.20\,\text{m}\), \[ E=0.40+\frac12(50)(0.20)^2 \] \[ E=1.40\,\text{J} \] At \(x=-0.12\,\text{m}\), \[ U=0.40+\frac12(50)(0.12)^2 \] \[ U=0.76\,\text{J} \] Therefore, \[ K=E-U=1.40-0.76 \] \[ K=0.64\,\text{J} \] \(\textbf{Speed at the observation point:}\) Using \[ K=\frac12mv^2 \] we obtain \[ 0.64=\frac12(0.50)v^2 \] \[ v^2=2.56 \] \[ |v|=1.60\,\text{m s}^{-1} \] Only speed magnitude is determined because the direction of motion is not specified. \(\textbf{Work between opposite displacements:}\) Work done by the conservative restoring force is \[ W=U_i-U_f \] The endpoints have equal displacement magnitudes: \[ U(-0.12)=U(+0.12) \] Consequently, \[ W=0 \] The restoring force may do positive work during one part of the journey and negative work during another, but the net work is zero. \(\textbf{Physical interpretation:}\) An arbitrary additive constant in potential energy changes neither force nor oscillation dynamics. The negative gradient of the potential-energy graph, rather than its positive slope, determines restoring-force direction. Energy conservation provides speed without requiring the initial phase. Equal potential energies at opposite displacements also imply zero net restoring-force work between those positions, regardless of the path direction. \(\textbf{Final answer:}\) \[ \boxed{k=50\,\text{N m}^{-1},\qquad F(-0.12)=+6.0\,\text{N},\qquad |v|=1.60\,\text{m s}^{-1},\qquad W=0} \]

Explanation: \(\textbf{Recovering the initial displacement from energy division:}\) For ideal SHM, \[ \frac{U}{E}=\frac{x^2}{A^2} \] and \[ \frac{K}{E}=1-\frac{x^2}{A^2} \] Initially, \[ \frac{K_0}{E}=\frac14 \] Therefore, \[ \frac{x_0^2}{A^2}=\frac34 \] The displacement is specified to be positive: \[ x_0=+\frac{\sqrt3A}{2} \] \(\textbf{Selecting the phase from the velocity direction:}\) Represent the displacement as \[ x=A\cos(\omega t+\phi) \] The initial displacement gives \[ \cos\phi=\frac{\sqrt3}{2} \] Since the particle is moving towards the positive extreme, \[ v_0=-A\omega\sin\phi\gt 0 \] Thus, a suitable initial phase is \[ \phi=-\frac{\pi}{6} \] \(\textbf{Finding the first return to the same kinetic energy:}\) The particle first reaches the positive extreme, reverses direction, and returns towards equilibrium. It again has kinetic-energy fraction \(1/4\) when its displacement is \(+\sqrt3A/2\), but its velocity is now negative. The corresponding phase is \[ \theta=+\frac{\pi}{6} \] The phase advance from the initial state is \[ \Delta\theta=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right) \] \[ \Delta\theta=\frac{\pi}{3} \] Using \(\omega=2\pi/T\), \[ \Delta t=\frac{\Delta\theta}{\omega} \] \[ \Delta t=\frac{\pi/3}{2\pi/T} \] \[ \Delta t=\frac T6 \] \(\textbf{Energy division after one-quarter period:}\) At \(t=T/4\), the phase is \[ \theta=-\frac{\pi}{6}+\frac{\pi}{2} \] \[ \theta=\frac{\pi}{3} \] Thus, \[ x(T/4)=A\cos\frac{\pi}{3} \] \[ x(T/4)=\frac A2 \] The potential-energy fraction is \[ \frac UE=\frac{x^2}{A^2}=\frac14 \] The remaining fraction is kinetic: \[ \frac KE=\frac34 \] Therefore, \[ \frac KU=3 \] \(\textbf{Physical interpretation:}\) Kinetic energy is determined by the square of velocity and does not distinguish motion towards or away from a turning point. The initial velocity direction therefore supplies essential phase information. The first repetition of an energy value can occur much earlier than one full displacement period. After a quarter period, the particle has moved from near the positive extreme to a smaller positive displacement, transferring potential energy into kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{x_0=+\frac{\sqrt3A}{2},\qquad t_{\mathrm{return}}=\frac T6,\qquad\left.\frac KU\right|_{t=T/4}=3} \]

Explanation: \(\textbf{Determining amplitude from the zero-kinetic-energy observation:}\) For an ideal spring-mass oscillator, \[ K=E-\frac12kx^2 \] The kinetic energy vanishes at a turning point. Since the observation at \(x=-0.20\,\text{m}\) has \(K=0\), \[ A=0.20\,\text{m} \] The total mechanical energy is therefore \[ E=\frac12k(0.20)^2 \] \(\textbf{Using the second energy measurement to recover stiffness:}\) At \(x=+0.10\,\text{m}\), \[ K=\frac12k(A^2-x^2) \] Substituting, \[ 0.75=\frac12k[(0.20)^2-(0.10)^2] \] \[ 0.75=\frac12k(0.030) \] \[ k=50\,\text{N m}^{-1} \] \(\textbf{Total mechanical energy:}\) The oscillator's total energy is \[ E=\frac12kA^2 \] \[ E=\frac12(50)(0.20)^2 \] \[ E=1.0\,\text{J} \] At the first observation, the potential energy is \(0.25\,\text{J}\), leaving the measured \(0.75\,\text{J}\) as kinetic energy. \(\textbf{Recovering mass from the measured speed:}\) At \(x=+0.10\,\text{m}\), \[ K=\frac12mv^2 \] Therefore, \[ 0.75=\frac12m(1.50)^2 \] \[ m=\frac{1.50}{2.25} \] \[ m=\frac23\,\text{kg} \] \(\textbf{Maximum speed at equilibrium:}\) At equilibrium, all the oscillation energy is kinetic: \[ E=\frac12mv_{\max}^2 \] Thus, \[ v_{\max}=\sqrt{\frac{2E}{m}} \] \[ v_{\max}=\sqrt{\frac{2(1.0)}{2/3}} \] \[ v_{\max}=\sqrt3\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) A zero-kinetic-energy observation identifies a turning point without requiring any timing information. A second observation at a different displacement makes it possible to recover stiffness from the reduction of kinetic energy away from equilibrium. The measured particle speed then determines mass independently. The largest speed occurs at equilibrium because spring potential energy is smallest there. The signed displacements in the table do not imply signed kinetic energies; kinetic energy remains nonnegative. \(\textbf{Final answer:}\) \[ \boxed{k=50\,\text{N m}^{-1},\qquad m=\frac23\,\text{kg},\qquad E=1.0\,\text{J},\qquad v_{\max}=\sqrt3\,\text{m s}^{-1}} \]

Explanation: \(\textbf{Energy fractions along the oscillator's motion:}\) Because the particle begins at its positive extreme, \[ x=A\cos(\omega t) \] where \[ \omega=\frac{2\pi}{T} \] The potential-energy fraction is \[ \frac UE=\cos^2(\omega t) \] The kinetic-energy fraction is \[ \frac KE=\sin^2(\omega t) \] \(\textbf{Converting the first energy inequality:}\) The condition is \[ K\gt 3U \] Using \(E=K+U\), this becomes \[ E-U\gt 3U \] \[ U\lt \frac E4 \] Therefore, \[ \cos^2(\omega t)\lt \frac14 \] \[ |\cos(\omega t)|\lt \frac12 \] \(\textbf{First time interval satisfying the condition:}\) During the first half period, the relevant phase interval is \[ \frac{\pi}{3}\lt \omega t\lt \frac{2\pi}{3} \] Converting phase into time, \[ \frac{\pi/3}{2\pi/T}\lt t\lt \frac{2\pi/3}{2\pi/T} \] \[ \frac T6\lt t\lt \frac T3 \] \(\textbf{Total duration over one period:}\) The first interval has duration \[ \frac T3-\frac T6=\frac T6 \] An identical-duration interval occurs during the second half period. Therefore, \[ \Delta t_{K\gt 3U}=2\left(\frac T6\right) \] \[ \Delta t_{K\gt 3U}=\frac T3 \] The fraction of one period is \[ \frac{\Delta t_{K\gt 3U}}{T}=\frac13 \] \(\textbf{Time for potential energy to dominate:}\) For \[ U\gt 3K \] we similarly obtain \[ U\gt \frac{3E}{4} \] Therefore, \[ \cos^2(\omega t)\gt \frac34 \] \[ |\cos(\omega t)|\gt \frac{\sqrt3}{2} \] These regions occur around the two turning-point phases. Their combined angular extent is \(2\pi/3\) within a complete \(2\pi\) phase cycle. Thus, \[ \frac{\Delta t_{U\gt 3K}}T=\frac13 \] \(\textbf{Physical interpretation:}\) The oscillator does not spend equal time in equal displacement intervals because its speed varies with position. Energy inequalities must therefore be converted into phase intervals before calculating time fractions. The condition \(K\gt 3U\) selects regions near equilibrium, where motion is relatively fast. The condition \(U\gt 3K\) selects regions near turning points, where speed is small. These two distinct conditions happen to occupy equal fractions of the period, although their spatial regions are different. \(\textbf{Final answer:}\) \[ \boxed{\frac T6\lt t\lt \frac T3,\qquad \frac{\Delta t_{K\gt 3U}}T=\frac13,\qquad \frac{\Delta t_{U\gt 3K}}T=\frac13} \]

Explanation: Potential energy is defined relative to a chosen reference level. Adding a constant \(C\) changes that reference without changing the physical spring or its force law. The restoring force is obtained from the negative displacement derivative of potential energy. Since the derivative of a constant is zero, both potential-energy descriptions give the same force \(-kx\). The work done by the spring between two positions equals the decrease in its potential energy. The added constant appears in both endpoint values and cancels when their difference is calculated. Thus, the work between any two positions remains unchanged. The numerical total mechanical energy in the second description is \(E_2=E_1+C\). However, kinetic energy at displacement \(x\) is obtained from \(K=E-U(x)\). In the second description, both \(E\) and \(U\) increase by \(C\), so their difference remains exactly the same. Consequently, the particle's kinetic energy and speed at each displacement are unchanged. The Assertion is therefore true. The Reason accurately identifies the two mathematical properties responsible for this invariance and is also true. It correctly explains why changing the reference level changes numerical energy values without changing measurable oscillator dynamics.

Explanation: \(\textbf{Work from the spring potential-energy difference:}\) For the ideal spring, \[ U(x)=\frac12kx^2 \] The spring's work between two positions is \[ W=U_i-U_f \] With \(x_i=+0.15\,\text{m}\) and \(x_f=-0.05\,\text{m}\), \[ W=\frac12(100)[(0.15)^2-(-0.05)^2] \] \[ W=50(0.0225-0.0025) \] \[ W=+1.0\,\text{J} \] \(\textbf{Change in kinetic energy:}\) The spring is the only force doing work along the smooth horizontal motion. By the work-energy theorem, \[ \Delta K=W \] Therefore, \[ K_f-K_i=+1.0\,\text{J} \] The particle has gained kinetic energy over the complete journey. \(\textbf{Final speed from conservation of energy:}\) The total mechanical energy is fixed by amplitude: \[ E=\frac12kA^2 \] \[ E=\frac12(100)(0.20)^2 \] \[ E=2.0\,\text{J} \] At the final position, \[ U_f=\frac12(100)(0.05)^2 \] \[ U_f=0.125\,\text{J} \] Therefore, \[ K_f=E-U_f \] \[ K_f=1.875\,\text{J} \] Using \[ K_f=\frac12mv_f^2 \] we obtain \[ 1.875=\frac12(1.0)v_f^2 \] \[ v_f^2=3.75 \] The particle is still moving towards negative \(x\): \[ v_f=-\frac{\sqrt{15}}2\,\text{m s}^{-1} \] \(\textbf{Identifying the power-transfer reversal:}\) The instantaneous power delivered by the spring is \[ P=Fv=-kxv \] From \(x=+0.15\,\text{m}\) to equilibrium, both the spring force and velocity point towards negative \(x\). Their product is positive, so kinetic energy increases. After the particle crosses into negative displacement, the spring force points towards positive \(x\) while velocity remains negative. The power then becomes negative and kinetic energy decreases. \(\textbf{Physical interpretation:}\) Positive net spring work does not mean the spring supplies positive power at every point along the journey. Power depends on the instantaneous directions of force and velocity, whereas net work depends on the initial and final potential energies. The particle speeds up while approaching equilibrium and slows down after passing it. Because its final displacement magnitude is smaller than its initial magnitude, the overall work remains positive despite the final interval of negative power. \(\textbf{Final answer:}\) \[ \boxed{W=+1.0\,\text{J},\qquad \Delta K=+1.0\,\text{J},\qquad v_f=-\frac{\sqrt{15}}2\,\text{m s}^{-1}} \]

Explanation: \(\textbf{Initial displacement and phase:}\) For SHM, \[ \frac UE=\frac{x^2}{A^2} \] The initial potential-energy fraction is \(3/4\). Therefore, \[ \frac{x_0^2}{A^2}=\frac34 \] The positive displacement gives \[ x_0=+\frac{\sqrt3A}{2} \] Choose \[ x=A\cos(\omega t+\phi) \] The particle is moving towards equilibrium, so its initial velocity is negative. Therefore, a compatible phase is \[ \phi=\frac{\pi}{6} \] \(\textbf{First equality of kinetic and potential energies:}\) The equality \(K=U\) implies \[ U=\frac E2 \] Thus, \[ \cos^2\theta=\frac12 \] As phase advances from \(\pi/6\), the first solution is \[ \theta=\frac{\pi}{4} \] The required phase advance is \[ \Delta\theta_1=\frac{\pi}{4}-\frac{\pi}{6} \] \[ \Delta\theta_1=\frac{\pi}{12} \] Therefore, \[ t_1=\frac{\Delta\theta_1}{2\pi/T} \] \[ t_1=\frac T{24} \] \(\textbf{First occurrence of }K=3U\textbf{:}\) For \[ K=3U \] the energy fractions are \[ U=\frac E4,\qquad K=\frac{3E}{4} \] Thus, \[ \cos^2\theta=\frac14 \] The first solution after the initial phase is \[ \theta=\frac{\pi}{3} \] Therefore, \[ \Delta\theta_2=\frac{\pi}{3}-\frac{\pi}{6} \] \[ \Delta\theta_2=\frac{\pi}{6} \] Hence, \[ t_2=\frac{\pi/6}{2\pi/T} \] \[ t_2=\frac T{12} \] \(\textbf{Next instant at which all energy is potential:}\) The condition \(U=E\) requires \[ |x|=A \] The particle initially moves towards equilibrium and then proceeds to the negative extreme. That extreme corresponds to phase \[ \theta=\pi \] The phase advance is \[ \Delta\theta_3=\pi-\frac{\pi}{6} \] \[ \Delta\theta_3=\frac{5\pi}{6} \] Thus, \[ t_3=\frac{5\pi/6}{2\pi/T} \] \[ t_3=\frac{5T}{12} \] \(\textbf{Physical interpretation:}\) An energy measurement determines displacement magnitude but not its sign or the direction of motion. The additional initial-state information resolves that ambiguity. As the particle approaches equilibrium, potential energy decreases and kinetic energy increases. After crossing equilibrium, potential energy rises again as the particle approaches the opposite turning point. The three requested events therefore occur at distinct phase positions that cannot be recovered simply by dividing the period into equal energy fractions. \(\textbf{Final answer:}\) \[ \boxed{t_{K=U}=\frac T{24},\qquad t_{K=3U}=\frac T{12},\qquad t_{U=E}=\frac{5T}{12}} \]

Explanation: \(\textbf{Static equilibrium of the vertical spring:}\) At equilibrium, the upward spring force balances weight: \[ kx_{\mathrm{eq}}=mg \] Therefore, \[ x_{\mathrm{eq}}=\frac{mg}{k} \] \[ x_{\mathrm{eq}}=\frac{1.0(10)}{100} \] \[ x_{\mathrm{eq}}=0.10\,\text{m} \] \(\textbf{Turning-point positions:}\) The oscillation amplitude about equilibrium is \(0.050\,\text{m}\). Thus, \[ x_{\mathrm{upper}}=0.10-0.050 \] \[ x_{\mathrm{upper}}=0.050\,\text{m} \] \[ x_{\mathrm{lower}}=0.10+0.050 \] \[ x_{\mathrm{lower}}=0.150\,\text{m} \] \(\textbf{Spring and gravitational energies at the upper extreme:}\) The spring potential energy is \[ U_s=\frac12kx^2 \] At \(x=0.050\,\text{m}\), \[ U_{s,\mathrm{upper}}=\frac12(100)(0.050)^2 \] \[ U_{s,\mathrm{upper}}=+0.125\,\text{J} \] With downward displacement positive, gravitational potential energy is \[ U_g=-mgx \] Therefore, \[ U_{g,\mathrm{upper}}=-1.0(10)(0.050) \] \[ U_{g,\mathrm{upper}}=-0.50\,\text{J} \] \(\textbf{Energies at the lower extreme:}\) At \(x=0.150\,\text{m}\), \[ U_{s,\mathrm{lower}}=\frac12(100)(0.150)^2 \] \[ U_{s,\mathrm{lower}}=1.125\,\text{J} \] The gravitational contribution is \[ U_{g,\mathrm{lower}}=-1.0(10)(0.150) \] \[ U_{g,\mathrm{lower}}=-1.50\,\text{J} \] \(\textbf{Conserved total mechanical energy:}\) At either turning point, kinetic energy is zero. Therefore, \[ E=U_s+U_g \] At the upper turning point, \[ E=0.125-0.50 \] \[ E=-0.375\,\text{J} \] At the lower turning point, \[ E=1.125-1.50 \] \[ E=-0.375\,\text{J} \] The two turning-point values agree. \(\textbf{Maximum speed at equilibrium:}\) The angular frequency is \[ \omega=\sqrt{\frac km} \] \[ \omega=10\,\text{rad s}^{-1} \] The maximum speed is \[ v_{\max}=\omega A \] \[ v_{\max}=10(0.050) \] \[ v_{\max}=0.50\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) A vertical spring stores elastic energy while gravity contributes a separate potential energy. As the mass moves between turning points, the two potential-energy contributions change by different amounts but their sum is the same at both extremes. The total energy can be negative because the selected potential-energy reference is arbitrary. This does not mean that kinetic energy is negative or that mechanical energy is not conserved. When displacement is measured from the shifted equilibrium and the potential-energy minimum is chosen as zero, the familiar positive SHM energy \(\frac12kA^2\) is recovered. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} E=-0.375\,\text{J},\qquad (U_s,U_g)_{\mathrm{upper}}=(0.125,-0.50)\,\text{J}\\ (U_s,U_g)_{\mathrm{lower}}=(1.125,-1.50)\,\text{J},\qquad v_{\max}=0.50\,\text{m s}^{-1} \end{gathered}} \]

Explanation: \(\textbf{Energy equality and amplitude adjustment:}\) The conserved oscillation energy of an ideal spring-mass system is \[ E=\frac12kA^2 \] For the first oscillator, \[ E_1=\frac12(50)(0.12)^2 \] \[ E_1=0.36\,\text{J} \] The second spring constant is \[ k_2=4(50)=200\,\text{N m}^{-1} \] Equal oscillation energies require \[ \frac12k_2A_2^2=0.36 \] Therefore, \[ A_2^2=\frac{0.72}{200} \] \[ A_2=0.060\,\text{m} \] \(\textbf{Comparing maximum particle speeds:}\) The maximum speed is \[ v_{\max}=\omega A \] Since \[ \omega=\sqrt{\frac km} \] we have \[ v_{\max}=A\sqrt{\frac km} \] For equal mass and equal total energy, \[ \frac12mv_{\max}^2=E \] Thus, both maximum speeds are equal: \[ \frac{v_{\max,2}}{v_{\max,1}}=1 \] \(\textbf{Comparing maximum accelerations:}\) The maximum acceleration is \[ a_{\max}=\omega^2A=\frac{kA}{m} \] Hence, \[ \frac{a_{\max,2}}{a_{\max,1}} =\frac{k_2A_2}{k_1A_1} \] Substituting, \[ \frac{a_{\max,2}}{a_{\max,1}} =\frac{200(0.060)}{50(0.12)} \] \[ \frac{a_{\max,2}}{a_{\max,1}}=2 \] \(\textbf{Kinetic energies at the common laboratory displacement:}\) At \(x=+0.060\,\text{m}\), the first oscillator has potential energy \[ U_1=\frac12(50)(0.060)^2 \] \[ U_1=0.09\,\text{J} \] Therefore, \[ K_1=0.36-0.09 \] \[ K_1=0.27\,\text{J} \] For the second oscillator, \(x=+0.060\,\text{m}\) equals its amplitude. It is at a turning point, so \[ K_2=0 \] \(\textbf{Physical interpretation:}\) Holding total energy constant while increasing stiffness requires amplitude to decrease. The maximum speed remains unchanged because total energy and mass are unchanged. Maximum acceleration nevertheless increases because the stiffer restoring force acts over the reduced amplitude. The same numerical displacement can represent an interior point of one oscillator's trajectory and a turning point of another. Consequently, equal total energies do not imply equal kinetic energies at the same position. \(\textbf{Final answer:}\) \[ \boxed{A_2=0.060\,\text{m},\qquad \frac{v_{\max,2}}{v_{\max,1}}=1,\qquad \frac{a_{\max,2}}{a_{\max,1}}=2,\qquad K_1=0.27\,\text{J},\quad K_2=0} \]

Explanation: \(\textbf{Constructing displacement and velocity functions:}\) The mass crosses equilibrium towards positive \(x\) at \(t=0\). A compatible displacement function is \[ x=A\sin(\omega t) \] Therefore, \[ v=A\omega\cos(\omega t) \] With the given parameters, \[ x=0.10\sin(20t) \] \[ v=2.0\cos(20t)\,\text{m s}^{-1} \] \(\textbf{Instantaneous spring power:}\) The spring force is \[ F=-kx \] The power delivered by the spring is \[ P_s=Fv=-kxv \] Substituting the SHM functions, \[ P_s=-kA^2\omega\sin(\omega t)\cos(\omega t) \] Use \[ 2\sin\theta\cos\theta=\sin2\theta \] to obtain \[ P_s=-\frac{kA^2\omega}{2}\sin(2\omega t) \] \(\textbf{Power limits:}\) The maximum magnitude is \[ P_{\max}=\frac{kA^2\omega}{2} \] Substituting, \[ P_{\max}=\frac{80(0.10)^2(20)}2 \] \[ P_{\max}=8.0\,\text{W} \] Thus, \[ -8.0\leq P_s\leq+8.0\,\text{W} \] \(\textbf{First instant of greatest positive power:}\) The power is positive and maximum when \[ -\sin(40t)=+1 \] The first positive solution satisfies \[ 40t=\frac{3\pi}{2} \] Therefore, \[ t=\frac{3\pi}{80}\,\text{s} \] \(\textbf{Particle state at that instant:}\) At this time, \[ 20t=\frac{3\pi}{4} \] The displacement is \[ x=A\sin\frac{3\pi}{4} \] \[ x=+\frac A{\sqrt2} \] The velocity is \[ v=A\omega\cos\frac{3\pi}{4} \] \[ v=-\frac{A\omega}{\sqrt2} \] The mass is moving towards negative \(x\). \(\textbf{Energy-transfer interpretation:}\) At the specified event, displacement is positive, so the spring force points towards negative \(x\). Velocity also points towards negative \(x\). Their product is positive because the spring transfers potential energy into kinetic energy. The maximum positive power occurs neither at equilibrium nor at the turning point. At equilibrium the spring force vanishes, while at the turning point velocity vanishes. The greatest transfer rate occurs at an intermediate displacement where the magnitudes of both force and velocity are substantial. \(\textbf{Final answer:}\) \[ \boxed{P_{s,\max}=+8.0\,\text{W},\qquad P_{s,\min}=-8.0\,\text{W},\qquad t_{\mathrm{first\ positive\ max}}=\frac{3\pi}{80}\,\text{s},\qquad x=+\frac A{\sqrt2},\quad v\lt 0} \]

Explanation: For ideal SHM, the spring potential energy depends on the square of displacement. All four states have displacement magnitude \(A/2\), so each has potential energy \(E/4\). Their kinetic energies are therefore all \(3E/4\), making Statement I correct. The instantaneous spring force is \(-kx\), while its power is the product \(Fv\). In P, displacement and velocity are both positive, so the restoring force opposes motion and power is negative. In S, displacement and velocity are both negative; the restoring force is positive, again opposing motion. Statement II is correct. In Q, the mass has positive displacement but moves towards negative \(x\), so both force and velocity point towards negative \(x\). The spring therefore supplies positive power and increases kinetic energy. In R, displacement is negative and velocity positive; the restoring force and velocity again have the same direction. Statement III is correct. Acceleration depends on signed displacement through \(a=-\omega^2x\), not on kinetic energy alone. The acceleration is negative in P and Q but positive in R and S, making Statement IV false. Although P and Q have equal displacement and energies, their velocities have opposite signs. They consequently move in opposite directions and occupy different phases of the oscillation. Statement V is false. The table demonstrates that energy measurements can fail to distinguish states that differ in displacement sign, velocity direction or acceleration direction. Complete instantaneous mechanical state requires more information than kinetic and potential energy values alone.

Explanation: \(\textbf{Pendulum potential energy at small angles:}\) The bob rises above its lowest position by \[ h=\ell(1-\cos\theta) \] For sufficiently small angles, \[ 1-\cos\theta\simeq\frac{\theta^2}{2} \] Therefore, \[ U(\theta)\simeq\frac12mg\ell\theta^2 \] The lowest position is the zero of potential energy. \(\textbf{Energy at the angular turning point:}\) At \(\theta=\theta_0\), the bob is momentarily stationary. Its entire oscillation energy is gravitational potential energy: \[ E\simeq\frac12mg\ell\theta_0^2 \] Substituting, \[ E=\frac12(0.20)(10)(1.00)(0.10)^2 \] \[ E=0.010\,\text{J} \] \(\textbf{Speed at the lowest point:}\) At the lowest point, gravitational potential energy is zero. Conservation of mechanical energy gives \[ \frac12mv_{\max}^2=E \] Therefore, \[ v_{\max}=\sqrt{\frac{2E}{m}} \] \[ v_{\max}=\sqrt{\frac{2(0.010)}{0.20}} \] \[ v_{\max}=\frac1{\sqrt{10}}\,\text{m s}^{-1} \] \(\textbf{Energy partition at half the angular amplitude:}\) For small-angle pendulum SHM, \[ \frac UE=\frac{\theta^2}{\theta_0^2} \] At \[ |\theta|=\frac{\theta_0}{2} \] the potential-energy fraction is \[ \frac UE=\frac14 \] The kinetic fraction is \[ \frac KE=1-\frac14 \] \[ \frac KE=\frac34 \] Thus, \[ K:U=3:1 \] \(\textbf{Physical interpretation:}\) The pendulum's restoring potential energy is gravitational, unlike the elastic potential energy of a horizontal spring. For sufficiently small angular oscillations, however, both potentials have the same quadratic dependence on displacement from equilibrium. Consequently, the same energy-fraction rules apply when angular displacement is measured relative to angular amplitude. The mass appears in total energy but cancels from the maximum-speed expression derived from gravitational energy. At half the angular amplitude, the pendulum already has three-fourths of its total energy in kinetic form. \(\textbf{Final answer:}\) \[ \boxed{E=0.010\,\text{J},\qquad v_{\max}=\frac1{\sqrt{10}}\,\text{m s}^{-1},\qquad K:U=3:1} \]

Explanation: \(\textbf{Relating kinetic energy to squared restoring force:}\) For SHM, \[ E=K+U \] The spring potential energy is \[ U=\frac12kx^2 \] Since restoring-force magnitude is \(F=k|x|\), \[ x^2=\frac{F^2}{k^2} \] Therefore, \[ U=\frac{F^2}{2k} \] Consequently, \[ K=E-\frac{F^2}{2k} \] This predicts a straight-line graph of \(K\) against \(F^2\). \(\textbf{Spring stiffness from the graph slope:}\) The graph slope is \[ S=-\frac1{2k} \] The measured value is \[ S=-0.0050\,\text{J N}^{-2} \] Therefore, \[ \frac1{2k}=0.0050 \] \[ k=100\,\text{N m}^{-1} \] \(\textbf{Total energy from the vertical intercept:}\) At \(F=0\), the mass is at equilibrium. Potential energy is zero, so kinetic energy equals total mechanical energy. The vertical intercept therefore gives \[ E=1.80\,\text{J} \] \(\textbf{Recovering particle mass from the measured period:}\) The period is \[ T=\frac{2\pi}{5}\,\text{s} \] Thus, \[ \omega=\frac{2\pi}{T} \] \[ \omega=5\,\text{rad s}^{-1} \] For a spring-mass oscillator, \[ \omega^2=\frac km \] Therefore, \[ m=\frac{k}{\omega^2} \] \[ m=\frac{100}{25} \] \[ m=4.0\,\text{kg} \] \(\textbf{Amplitude from the recovered total energy:}\) \[ E=\frac12kA^2 \] Therefore, \[ A=\sqrt{\frac{2E}{k}} \] \[ A=\sqrt{\frac{3.60}{100}} \] \[ A=\sqrt{0.036} \] \[ A=\frac3{5\sqrt{10}}\,\text{m} \] \(\textbf{Maximum force at the turning points:}\) The greatest force magnitude is \[ F_{\max}=kA \] \[ F_{\max}=100\left(\frac3{5\sqrt{10}}\right) \] \[ F_{\max}=6\sqrt{10}\,\text{N} \] \(\textbf{Physical interpretation:}\) The kinetic-energy graph is linear in \(F^2\), not in signed force or displacement. Its slope encodes spring stiffness, while its intercept encodes total oscillation energy. The independently measured period then identifies mass. Combining this information recovers amplitude and maximum restoring-force magnitude. The graph also reflects the fact that kinetic energy is largest where spring force vanishes and zero where restoring-force magnitude is greatest. \(\textbf{Final answer:}\) \[ \boxed{k=100\,\text{N m}^{-1},\qquad m=4.0\,\text{kg},\qquad A=\frac3{5\sqrt{10}}\,\text{m},\qquad F_{\max}=6\sqrt{10}\,\text{N}} \]

Explanation: \(\textbf{Distinguishing kinetic-energy period from displacement period:}\) An SHM particle reaches maximum speed whenever it crosses equilibrium. It crosses equilibrium twice in every complete oscillation. Consequently, consecutive kinetic-energy maxima are separated by \[ \Delta t_K=\frac T2 \] The measured separation is \[ \Delta t_K=0.120-0.040 \] \[ \Delta t_K=0.080\,\text{s} \] Therefore, \[ T=2(0.080) \] \[ T=0.160\,\text{s} \] \(\textbf{Maximum speed from total mechanical energy:}\) At equilibrium, all the oscillator's energy is kinetic: \[ E=\frac12mv_{\max}^2 \] Therefore, \[ v_{\max}=\sqrt{\frac{2E}{m}} \] Substituting, \[ v_{\max}=\sqrt{\frac{2(0.80)}{0.40}} \] \[ v_{\max}=2.0\,\text{m s}^{-1} \] \(\textbf{Displacement phase at the specified instant:}\) Because the particle begins at its positive extreme, \[ x=A\cos(\omega t) \] The angular frequency is \[ \omega=\frac{2\pi}{T} \] At \(t=0.060\,\text{s}\), \[ \omega t=\frac{2\pi}{0.160}(0.060) \] \[ \omega t=\frac{3\pi}{4} \] Therefore, \[ x(0.060)=A\cos\frac{3\pi}{4} \] \[ x(0.060)=-\frac A{\sqrt2} \] \(\textbf{Kinetic energy at the observation instant:}\) The potential-energy fraction is \[ \frac UE=\frac{x^2}{A^2} \] Thus, \[ \frac UE=\frac12 \] The kinetic fraction is also one-half: \[ K=\frac E2 \] \[ K=0.40\,\text{J} \] \(\textbf{Determining the direction of energy transfer:}\) At phase \(3\pi/4\), displacement is negative. The restoring force therefore points towards positive \(x\). The velocity follows from \[ v=-A\omega\sin(\omega t) \] Since \[ \sin\frac{3\pi}{4}\gt 0 \] the velocity is negative. Force and velocity are opposite, so \[ P=Fv\lt 0 \] The restoring force is removing kinetic energy as the particle travels towards the negative extreme. \(\textbf{Physical interpretation:}\) The energy sensor records two kinetic-energy maxima during one complete displacement cycle. Its peak-to-peak timing therefore gives half the displacement period, not the full period. At the specified observation, kinetic and potential energies are equal, but this alone does not establish whether kinetic energy is increasing or decreasing. The displacement phase and velocity direction show that the particle has already crossed equilibrium and is moving away from it. Consequently, the restoring force does negative work at that instant. \(\textbf{Final answer:}\) \[ \boxed{T=0.160\,\text{s},\qquad v_{\max}=2.0\,\text{m s}^{-1},\qquad K(0.060)=0.40\,\text{J},\qquad P\lt 0} \]