Oscillations and Waves MCQs with Answers – Part 3 (NEET Physics)

Explanation: \(\textbf{Initial energy distribution:}\) At the specified initial instant, \[ K_0=0.20\,\text{J} \] \[ U_0=0.60\,\text{J} \] The conserved total mechanical energy is \[ E=K_0+U_0 \] \[ E=0.80\,\text{J} \] \(\textbf{Quarter-period transformation of SHM:}\) For SHM, write \[ x=A\cos\theta \] and \[ v=-A\omega\sin\theta \] After one-quarter period, the phase advances by \(\pi/2\). Therefore, \[ x_1=A\cos\left(\theta+\frac{\pi}{2}\right) \] Using the trigonometric identity, \[ x_1=-A\sin\theta \] Since \(v_0=-A\omega\sin\theta\), \[ x_1=\frac{v_0}{\omega} \] \(\textbf{Velocity after one-quarter period:}\) Similarly, \[ v_1=-A\omega\sin\left(\theta+\frac{\pi}{2}\right) \] \[ v_1=-A\omega\cos\theta \] Therefore, \[ v_1=-\omega x_0 \] \(\textbf{Kinetic energy at the later instant:}\) \[ K_1=\frac12mv_1^2 \] Substituting \(v_1=-\omega x_0\), \[ K_1=\frac12m\omega^2x_0^2 \] But the initial potential energy is \[ U_0=\frac12m\omega^2x_0^2 \] Therefore, \[ K_1=U_0 \] \[ K_1=0.60\,\text{J} \] \(\textbf{Determining the later displacement sign:}\) The initial velocity is negative. Since \[ x_1=\frac{v_0}{\omega} \] and \(\omega\gt 0\), \[ x_1\lt 0 \] Thus, the particle lies on the negative side of equilibrium at the later instant. \(\textbf{Determining the later velocity sign:}\) The initial displacement is positive. Since \[ v_1=-\omega x_0 \] we obtain \[ v_1\lt 0 \] The particle therefore has negative displacement and negative velocity. It is moving away from equilibrium towards the negative extreme. \(\textbf{Direction of energy transfer:}\) At negative displacement, the restoring force is positive. It opposes the negative velocity. Therefore, the restoring force performs negative instantaneous power, and kinetic energy is decreasing. Potential energy is increasing by the same rate because total energy remains conserved. \(\textbf{Physical interpretation:}\) A quarter-period shift exchanges the numerical values of kinetic and potential energy in ideal SHM. However, energy values alone do not specify the particle's direction of motion. The initial displacement and velocity signs determine the later signs through the quarter-period phase transformation. At the later instant, kinetic energy is \(0.60\,\text{J}\), but it is already decreasing as the particle moves towards the negative extreme. \(\textbf{Final answer:}\) The kinetic energy is \(0.60\,\text{J}\); displacement is negative, and kinetic energy is decreasing.

Explanation: \(\textbf{Meaning of zero kinetic energy:}\) For a particle executing SHM, \[ K=\frac12mv^2 \] Kinetic energy becomes zero only when velocity is zero. These instants occur at the two extreme positions. Consecutive turning points are separated by half a period. \(\textbf{Determining the period from sensor readings:}\) The consecutive zero-energy readings are separated by \[ \Delta t=0.30\,\text{s} \] Therefore, \[ \frac{T}{2}=0.30\,\text{s} \] \[ T=0.60\,\text{s} \] \(\textbf{Energy variation from the initial extreme:}\) Since the particle starts at the positive extreme, \[ x=A\cos(\omega t) \] Its velocity is \[ v=-A\omega\sin(\omega t) \] Therefore, its kinetic energy is \[ K=E\sin^2(\omega t) \] where the constant total energy equals the maximum kinetic energy: \[ E=0.80\,\text{J} \] \(\textbf{First occurrence of the required energy:}\) The required kinetic energy is \[ K=0.60\,\text{J} \] Hence, \[ 0.60=0.80\sin^2(\omega t) \] \[ \sin^2(\omega t)=\frac34 \] The first positive phase satisfying this condition is \[ \omega t=\frac{\pi}{3} \] \(\textbf{Displacement sign at this phase:}\) \[ x=A\cos\left(\frac{\pi}{3}\right) \] \[ x=+\frac{A}{2} \] Thus, the displacement is positive. \(\textbf{Direction of motion:}\) At the same phase, \[ v=-A\omega\sin\left(\frac{\pi}{3}\right) \] \[ v=-\frac{\sqrt3A\omega}{2} \] The velocity is negative. Because the displacement is positive, the particle is moving towards equilibrium. \(\textbf{Physical interpretation:}\) The kinetic-energy signal has half the period of the displacement signal because kinetic energy depends on the square of velocity. The first rise to \(0.60\,\text{J}\) occurs before the first equilibrium crossing. An energy measurement alone cannot distinguish positive and negative displacement, but the initial state and the requirement of the first occurrence resolve that ambiguity. The sensor therefore provides both a period measurement and, when combined with the initial condition, information about the particle's position and motion direction. \(\textbf{Final answer:}\) The particle has positive displacement and moves towards equilibrium, with \[ \boxed{T=0.60\,\text{s}} \]

Explanation: \(\textbf{Work of a conservative restoring force:}\) For a conservative force, \[ W=-\Delta U \] Therefore, \[ W=U_i-U_f \] The initial and final displacements are \[ x_i=\frac{3A}{4} \] \[ x_f=-\frac{A}{4} \] \(\textbf{Potential energy at the initial position:}\) \[ U_i=U_0+\frac12k\left(\frac{3A}{4}\right)^2 \] \[ U_i=U_0+\frac{9kA^2}{32} \] \(\textbf{Potential energy at the final position:}\) \[ U_f=U_0+\frac12k\left(-\frac{A}{4}\right)^2 \] \[ U_f=U_0+\frac{kA^2}{32} \] \(\textbf{Net work over the journey:}\) \[ W=U_i-U_f \] Substituting, \[ W=\left(U_0+\frac{9kA^2}{32}\right)-\left(U_0+\frac{kA^2}{32}\right) \] The arbitrary reference constant cancels. \[ W=\frac{8kA^2}{32} \] \[ W=+\frac{kA^2}{4} \] \(\textbf{Kinetic-energy change:}\) The work-energy theorem gives \[ \Delta K=W \] Therefore, \[ \Delta K=+\frac{kA^2}{4} \] \(\textbf{Understanding the force reversal:}\) During motion from \(+3A/4\) towards equilibrium, the restoring force acts in the direction of motion and does positive work. After the particle crosses equilibrium, the restoring force points towards positive displacement while the particle continues negatively. The restoring force then does negative work. However, the final displacement magnitude is smaller than the initial displacement magnitude. Consequently, the positive contribution exceeds the negative contribution. \(\textbf{Physical interpretation:}\) The work performed by a conservative restoring force depends on the initial and final positions, not on the detailed intermediate path. Crossing equilibrium does not make the net work zero. Changing the reference level \(U_0\) shifts both potential-energy values equally and cannot change their difference. The positive work signifies a net increase in the particle's kinetic energy over the specified journey. \(\textbf{Final answer:}\) \[ \boxed{W=\Delta K=+\frac{kA^2}{4}} \]

Explanation: \(\textbf{Angular frequency of the original oscillator:}\) Given, \[ m=1\,\text{kg},\qquad k=100\,\text{N m}^{-1} \] Therefore, \[ \omega=\sqrt{\frac{k}{m}} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Velocity immediately before the impulse:}\) The original amplitude is \[ A=0.10\,\text{m} \] At the specified instant, \[ x=+0.06\,\text{m} \] The SHM speed relation is \[ v_i^2=\omega^2(A^2-x^2) \] Substituting, \[ v_i^2=100(0.10^2-0.06^2) \] \[ v_i^2=100(0.0100-0.0036) \] \[ v_i^2=0.64\,\text{m}^2\text{s}^{-2} \] The velocity is directed negatively. \[ v_i=-0.80\,\text{m s}^{-1} \] \(\textbf{Velocity immediately after the impulse:}\) The impulse produces a signed velocity change \[ \Delta v=+0.50\,\text{m s}^{-1} \] Therefore, \[ v_f=v_i+\Delta v \] \[ v_f=-0.80+0.50 \] \[ v_f=-0.30\,\text{m s}^{-1} \] The block continues moving negatively, but more slowly. \(\textbf{Work performed by the impulse:}\) The block's position does not change during the instantaneous impulse, so its spring potential energy is unchanged. The impulse work equals the change in kinetic energy. \[ W_{\mathrm{imp}}=\frac12m(v_f^2-v_i^2) \] Substituting, \[ W_{\mathrm{imp}}=\frac12(1)[(-0.30)^2-(-0.80)^2] \] \[ W_{\mathrm{imp}}=\frac12(0.09-0.64) \] \[ W_{\mathrm{imp}}=-0.275\,\text{J} \] \(\textbf{Amplitude of the new free motion:}\) The spring and mass remain unchanged. Therefore, the angular frequency remains \(10\,\text{rad s}^{-1}\). Using the instantaneous state after the impulse, \[ A'^2=x^2+\frac{v_f^2}{\omega^2} \] Substituting, \[ A'^2=(0.06)^2+\frac{(-0.30)^2}{100} \] \[ A'^2=0.0036+0.0009 \] \[ A'^2=0.0045\,\text{m}^2 \] Therefore, \[ A'=\sqrt{45}\,\text{cm} \] \(\textbf{Energy verification:}\) The original mechanical energy is \[ E_i=\frac12kA^2 \] \[ E_i=0.50\,\text{J} \] After the impulse, \[ E_f=E_i+W_{\mathrm{imp}} \] \[ E_f=0.50-0.275 \] \[ E_f=0.225\,\text{J} \] Also, \[ \frac12kA'^2=50(0.0045)=0.225\,\text{J} \] The energy balance agrees with the new amplitude. \(\textbf{Physical interpretation:}\) A positive impulse need not do positive work. Here, the impulse acts opposite to the original negative velocity and removes kinetic energy. The spring's potential energy remains unchanged during the impulse because position is continuous. The reduced total mechanical energy produces a smaller subsequent amplitude without changing the natural frequency. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{imp}}=-0.275\,\text{J},\qquad A'=\sqrt{45}\,\text{cm}} \]

Explanation: \(\textbf{Identifying the equilibrium from the potential energy:}\) The conservative force is related to potential energy by \[ F=-\frac{dU}{dX} \] Differentiating the supplied function, \[ F=-(100X-10) \] \[ F=10-100X \] At equilibrium, \[ F=0 \] Therefore, \[ 10-100X_0=0 \] \[ X_0=+0.10\,\text{m} \] \(\textbf{Expressing potential energy about equilibrium:}\) Complete the square in the energy expression. \[ U(X)=3+50X^2-10X \] \[ U(X)=3+50(X^2-0.20X) \] \[ U(X)=3+50[(X-0.10)^2-0.01] \] Therefore, \[ U(X)=2.50+50(X-0.10)^2 \] \(\textbf{Minimum potential energy:}\) The quadratic term is nonnegative. Hence, the minimum potential energy occurs at \(X=X_0\). \[ U_{\min}=2.50\,\text{J} \] \(\textbf{Energy available for oscillation:}\) The total mechanical energy is \[ E=3.00\,\text{J} \] At equilibrium, the kinetic energy is maximum. \[ K_{\max}=E-U_{\min} \] \[ K_{\max}=3.00-2.50 \] \[ K_{\max}=0.50\,\text{J} \] \(\textbf{Turning-point coordinates:}\) At either turning point, \[ K=0 \] Therefore, \[ U(X)=E \] Substituting, \[ 2.50+50(X-0.10)^2=3.00 \] \[ 50(X-0.10)^2=0.50 \] \[ (X-0.10)^2=0.0100 \] Hence, \[ X-0.10=\pm0.10 \] The turning coordinates are \[ X_1=0 \] \[ X_2=+0.20\,\text{m} \] \(\textbf{Maximum speed:}\) At equilibrium, \[ K_{\max}=\frac12mv_{\max}^2 \] Substituting \(m=1\,\text{kg}\), \[ 0.50=\frac12(1)v_{\max}^2 \] \[ v_{\max}=1\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) The potential-energy function contains a term linear in the absolute coordinate, so equilibrium is shifted away from the origin. Completing the square reveals a quadratic potential well centred at \(X=0.10\,\text{m}\). The additive energy constant changes the numerical potential-energy reference but not the restoring force. Only the difference between total energy and minimum potential energy is available as kinetic energy at equilibrium. \(\textbf{Final answer:}\) \[ \boxed{X_0=+0.10\,\text{m},\qquad X_{\mathrm{turn}}=0,\ +0.20\,\text{m},\qquad v_{\max}=1\,\text{m s}^{-1}} \]

Explanation: \(\textbf{Locating static equilibrium:}\) For a vertical spring-block system, \[ kl_0=mg \] Therefore, \[ l_0=\frac{mg}{k} \] Substituting, \[ l_0=\frac{0.50(10)}{100} \] \[ l_0=0.05\,\text{m} \] The equilibrium position is \(5\,\text{cm}\) below the natural-length position. \(\textbf{Work performed by gravity:}\) The block moves downwards by \[ s=0.05\,\text{m} \] Gravity also acts downwards. Therefore, \[ W_g=mgs \] Substituting, \[ W_g=0.50(10)(0.05) \] \[ W_g=+0.25\,\text{J} \] \(\textbf{Work performed by the spring:}\) The spring is initially undeformed and becomes extended by \(0.05\,\text{m}\). Its force opposes the downward displacement. The work done by the spring is the negative change in elastic potential energy. \[ W_s=-\Delta U_s \] \[ W_s=-\frac12k(l_0^2-0^2) \] Substituting, \[ W_s=-\frac12(100)(0.05)^2 \] \[ W_s=-50(0.0025) \] \[ W_s=-0.125\,\text{J} \] \(\textbf{Net work and kinetic energy:}\) The block starts from rest. Therefore, \[ K_i=0 \] The work-energy theorem gives \[ K_f-K_i=W_g+W_s \] Hence, \[ K_f=0.25-0.125 \] \[ K_f=0.125\,\text{J} \] \(\textbf{Checking the motion at equilibrium:}\) The block is released from rest at the spring's natural length, which is \(5\,\text{cm}\) above equilibrium. It therefore begins at an extreme of its SHM about the equilibrium position. During its first descent to equilibrium, its speed increases to its maximum value. The net force becomes zero exactly at equilibrium, although its kinetic energy is then maximum. \(\textbf{Physical interpretation:}\) Gravity does positive work while the spring does negative work during the downward journey. The block gains kinetic energy because the positive gravitational work exceeds the energy stored in the spring. Using only the spring's elastic potential energy would give an incomplete energy balance for vertical motion. The total gravitational-plus-elastic potential energy decreases by \(0.125\,\text{J}\), exactly matching the kinetic-energy gain. \(\textbf{Final answer:}\) \[ \boxed{W_g=+0.25\,\text{J},\qquad W_s=-0.125\,\text{J},\qquad K_f=0.125\,\text{J}} \]

Explanation: The displacement of the oscillator is \(x=A\cos(\omega t)\), whose fundamental period is \(T=2\pi/\omega\). Differentiating gives \(v=-A\omega\sin(\omega t)\). Therefore, kinetic energy varies as \(K(t)=\frac12mA^2\omega^2\sin^2(\omega t)\). Potential energy, measured from zero at equilibrium, varies as \(U(t)=\frac12mA^2\omega^2\cos^2(\omega t)\). Both squared trigonometric functions repeat when their arguments advance by \(\pi\), instead of requiring the full \(2\pi\) phase advance of displacement. Hence, both nonconstant energy signals have fundamental period \(T/2\). Their maxima occur alternately: maximum kinetic energy coincides with minimum potential energy, and vice versa. Adding the two energies gives \(E=\frac12mA^2\omega^2[\sin^2(\omega t)+\cos^2(\omega t)]\). The trigonometric identity \(\sin^2\theta+\cos^2\theta=1\) makes the total mechanical energy constant. Strictly speaking, a constant signal has no unique fundamental period; it is invariant under every time translation. This does not affect the Assertion because only the kinetic and potential signals are assigned a fundamental period. The Reason is correct and explains both the halved energy periods and the constant total energy. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: \(\textbf{Restoring-force constant:}\) The mass and angular frequency are \[ m=0.50\,\text{kg},\qquad \omega=10\,\text{rad s}^{-1} \] Therefore, \[ k=m\omega^2 \] \[ k=0.50(10)^2 \] \[ k=50\,\text{N m}^{-1} \] \(\textbf{Amplitude from total mechanical energy:}\) For ideal SHM, \[ E=\frac12kA^2 \] Substituting, \[ 0.25=\frac12(50)A^2 \] \[ A^2=0.0100\,\text{m}^2 \] \[ A=0.10\,\text{m} \] \(\textbf{Displacement from the potential-energy fraction:}\) The condition is \[ U=\frac{E}{4} \] Using \[ \frac{U}{E}=\frac{x^2}{A^2} \] we obtain \[ \frac{x^2}{A^2}=\frac14 \] Therefore, \[ |x|=\frac{A}{2} \] The displacement is specified to be negative. \[ x=-0.05\,\text{m} \] \(\textbf{Signed velocity from kinetic energy:}\) The kinetic energy is \[ K=E-U \] \[ K=\frac{3E}{4} \] Substituting, \[ K=\frac34(0.25) \] \[ K=0.1875\,\text{J} \] Using \[ K=\frac12mv^2 \] we obtain \[ 0.1875=\frac12(0.50)v^2 \] \[ v^2=0.75\,\text{m}^2\text{s}^{-2} \] Since the velocity is positive, \[ v=+\frac{\sqrt3}{2}\,\text{m s}^{-1} \] \(\textbf{Instantaneous restoring force:}\) \[ F=-kx \] Substituting, \[ F=-50(-0.05) \] \[ F=+2.5\,\text{N} \] \(\textbf{Instantaneous power:}\) The power delivered by the restoring force is \[ P=Fv \] Substituting, \[ P=2.5\left(\frac{\sqrt3}{2}\right) \] \[ P=+\frac{5\sqrt3}{4}\,\text{W} \] \(\textbf{Direction of energy conversion:}\) The restoring force and velocity are both positive. Therefore, the force performs positive work and kinetic energy is increasing. Since total energy is conserved, potential energy must decrease at the same rate. The particle is moving towards equilibrium from the negative side. \(\textbf{Physical interpretation:}\) The potential-energy fraction determines displacement magnitude but not its sign. The supplied position and velocity directions are required to determine the signed restoring force and the sign of power. Positive restoring-force power means energy is being transferred from potential energy to kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{v=+\frac{\sqrt3}{2}\,\text{m s}^{-1},\qquad P=+\frac{5\sqrt3}{4}\,\text{W}} \] Potential energy is decreasing.

Explanation: \(\textbf{Effective stiffness of the series combination:}\) For two springs in series, \[ \frac1{k_{\mathrm{eff}}}=\frac1{k_1}+\frac1{k_2} \] Therefore, \[ k_{\mathrm{eff}}=\frac{k_1k_2}{k_1+k_2} \] Substituting, \[ k_{\mathrm{eff}}=\frac{100(300)}{100+300} \] \[ k_{\mathrm{eff}}=75\,\text{N m}^{-1} \] \(\textbf{Common force during the initial extension:}\) The total extension is \[ x=0.08\,\text{m} \] The series springs transmit the same force magnitude. \[ F=k_{\mathrm{eff}}x \] \[ F=75(0.08) \] \[ F=6\,\text{N} \] \(\textbf{Extension of the softer spring:}\) \[ x_1=\frac{F}{k_1} \] \[ x_1=\frac6{100} \] \[ x_1=0.06\,\text{m} \] \(\textbf{Extension of the stiffer spring:}\) \[ x_2=\frac{F}{k_2} \] \[ x_2=\frac6{300} \] \[ x_2=0.02\,\text{m} \] The extensions add to the specified \(8\,\text{cm}\). \(\textbf{Energy stored in the softer spring:}\) \[ U_1=\frac12k_1x_1^2 \] \[ U_1=\frac12(100)(0.06)^2 \] \[ U_1=0.18\,\text{J} \] \(\textbf{Energy stored in the stiffer spring:}\) \[ U_2=\frac12k_2x_2^2 \] \[ U_2=\frac12(300)(0.02)^2 \] \[ U_2=0.06\,\text{J} \] Therefore, \[ U_1:U_2=0.18:0.06 \] \[ U_1:U_2=3:1 \] \(\textbf{Total energy and maximum speed:}\) The block starts from rest, so its initial mechanical energy is entirely elastic. \[ E=U_1+U_2 \] \[ E=0.18+0.06 \] \[ E=0.24\,\text{J} \] At equilibrium, this energy becomes kinetic. \[ \frac12mv_{\max}^2=E \] Substituting, \[ \frac12(0.75)v_{\max}^2=0.24 \] \[ v_{\max}^2=0.64\,\text{m}^2\text{s}^{-2} \] \[ v_{\max}=0.80\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) The two series springs experience the same force, not the same extension. The softer spring extends more and stores more energy under this common force. At release, the total elastic energy is the sum of the energies in the individual springs. The equivalent spring constant correctly reproduces this total energy, but it does not indicate how that energy is distributed between the two springs. \(\textbf{Final answer:}\) \[ \boxed{U_1:U_2=3:1,\qquad E=0.24\,\text{J},\qquad v_{\max}=0.80\,\text{m s}^{-1}} \]

Explanation: \(\textbf{Comparing the two spring-potential energies:}\) The block passes through the same coordinate during both observations. The spring and equilibrium position are unchanged. Therefore, its spring deformation has the same value on both occasions. Since \[ U_s=\frac12kx^2 \] the spring potential energies are equal. \[ \Delta U_s=0 \] \(\textbf{Kinetic energy at the first passage:}\) The mass is \[ m=0.50\,\text{kg} \] The first signed velocity is \[ v_1=+4\,\text{m s}^{-1} \] Hence, \[ K_1=\frac12mv_1^2 \] \[ K_1=\frac12(0.50)(4)^2 \] \[ K_1=4.00\,\text{J} \] \(\textbf{Kinetic energy at the second passage:}\) The second signed velocity is \[ v_2=-3\,\text{m s}^{-1} \] Therefore, \[ K_2=\frac12mv_2^2 \] \[ K_2=\frac12(0.50)(-3)^2 \] \[ K_2=2.25\,\text{J} \] \(\textbf{Change in mechanical energy:}\) The mechanical-energy change is \[ \Delta E=(K_2+U_2)-(K_1+U_1) \] Because \(U_2=U_1\), \[ \Delta E=K_2-K_1 \] \[ \Delta E=2.25-4.00 \] \[ \Delta E=-1.75\,\text{J} \] \(\textbf{Work done by dissipative forces:}\) The spring force is conservative. The change in the block-spring mechanical energy therefore equals the work done by nonconservative forces. \[ W_{\mathrm{nc}}=\Delta E \] \[ W_{\mathrm{nc}}=-1.75\,\text{J} \] \(\textbf{Average nonconservative power:}\) The elapsed time is \[ \Delta t=0.70-0.20 \] \[ \Delta t=0.50\,\text{s} \] Hence, \[ \overline P_{\mathrm{nc}}=\frac{W_{\mathrm{nc}}}{\Delta t} \] \[ \overline P_{\mathrm{nc}}=\frac{-1.75}{0.50} \] \[ \overline P_{\mathrm{nc}}=-3.50\,\text{W} \] \(\textbf{Testing the ideal-SHM assumption:}\) In ideal undamped SHM, total mechanical energy remains constant. At a fixed displacement magnitude, kinetic energy and speed magnitude must therefore be unchanged from one passage to another. Here, the speed magnitude falls from \(4\,\text{m s}^{-1}\) to \(3\,\text{m s}^{-1}\). The observations are consequently incompatible with ideal undamped SHM. \(\textbf{Physical interpretation:}\) The opposite velocity signs alone would not demonstrate energy loss; ideal SHM also visits a coordinate in opposite directions. The decisive evidence is the decrease in velocity magnitude at the same coordinate. The negative work and average power quantify the net mechanical energy removed between readings. They do not identify the precise dissipative mechanism without additional information. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{nc}}=-1.75\,\text{J},\qquad \overline P_{\mathrm{nc}}=-3.50\,\text{W}} \] The motion is incompatible with ideal undamped SHM.

Explanation: \(\textbf{Initial speed at the switching position:}\) The original oscillator satisfies \[ v^2=\omega^2(A^2-x^2) \] At \[ x_i=+\frac{A}{2} \] we have \[ v_i^2=\omega^2\left(A^2-\frac{A^2}{4}\right) \] \[ v_i^2=\frac{3A^2\omega^2}{4} \] The velocity is negative because the particle moves towards the negative extreme. \[ v_i=-\frac{\sqrt3A\omega}{2} \] \(\textbf{New equilibrium after the force is applied:}\) The new net force is \[ F_{\mathrm{net}}=F_0-kx \] At equilibrium, \[ F_0-kx_0=0 \] Therefore, \[ x_0=\frac{F_0}{k} \] Substituting \(F_0=kA/2\), \[ x_0=\frac{A}{2} \] \(\textbf{State relative to the new equilibrium:}\) At switching, the particle is already located at \[ x_i=\frac{A}{2} \] Thus, its displacement from the new equilibrium is \[ y_i=x_i-x_0 \] \[ y_i=0 \] Its instantaneous velocity remains unchanged because the force application is impulse-free. \[ v_i=-\frac{\sqrt3A\omega}{2} \] \(\textbf{New oscillation amplitude:}\) A constant applied force shifts equilibrium but does not change the restoring-force coefficient \(k\). Therefore, the angular frequency remains \(\omega\). The particle is at the new equilibrium immediately after switching, so its speed magnitude is the maximum speed of the subsequent motion. \[ A'\omega=|v_i| \] Therefore, \[ A'=\frac{|v_i|}{\omega} \] \[ A'=\frac{\sqrt3A}{2} \] \(\textbf{Oscillatory energy about the new equilibrium:}\) With potential energy referenced to the new equilibrium minimum, \[ E'=\frac12kA'^2 \] Substituting, \[ E'=\frac12k\left(\frac{\sqrt3A}{2}\right)^2 \] \[ E'=\frac12k\frac{3A^2}{4} \] \[ E'=\frac34\left(\frac12kA^2\right) \] Since the original total energy is \[ E=\frac12kA^2 \] we obtain \[ E'=\frac{3E}{4} \] \(\textbf{Energy-reference interpretation:}\) The external force does no finite work during the instantaneous switching because displacement is continuous. However, the potential-energy function governing the subsequent motion is different. Its minimum is now at \(x=A/2\), not at \(x=0\). The reported \(E'\) is the energy measured above this new minimum. It is therefore incorrect to equate \(E'\) with the original energy without accounting for the shifted potential reference. \(\textbf{Physical interpretation:}\) The particle's original position coincides exactly with the newly established equilibrium. Its unchanged negative velocity determines the new amplitude directly. The speed at that location is less than the maximum speed of its original oscillation, so the new amplitude and oscillatory energy are correspondingly smaller. \(\textbf{Final answer:}\) \[ \boxed{A'=\frac{\sqrt3A}{2},\qquad E'=\frac{3E}{4}} \]

Explanation: \(\textbf{Velocities of the two particles:}\) The first displacement is \[ x_P=0.08\cos(10t)\,\text{m} \] Therefore, \[ v_P=-0.80\sin(10t)\,\text{m s}^{-1} \] Writing \(B\) in metres, the second displacement is \[ x_Q=B\sin(10t) \] Thus, \[ v_Q=10B\cos(10t) \] \(\textbf{Kinetic energy of particle }P\textbf{:}\) \[ K_P=\frac12m_Pv_P^2 \] Substituting \(m_P=1\,\text{kg}\), \[ K_P=\frac12(1)(0.80)^2\sin^2(10t) \] \[ K_P=0.32\sin^2(10t)\,\text{J} \] \(\textbf{Kinetic energy of particle }Q\textbf{:}\) \[ K_Q=\frac12m_Qv_Q^2 \] Substituting \(m_Q=4\,\text{kg}\), \[ K_Q=\frac12(4)(10B)^2\cos^2(10t) \] \[ K_Q=200B^2\cos^2(10t)\,\text{J} \] where \(B\) is expressed in metres. \(\textbf{Condition for a time-independent kinetic-energy sum:}\) The measured sum is \[ K_P+K_Q=0.32\sin^2(10t)+200B^2\cos^2(10t) \] For this expression to remain constant at every phase, the coefficients of the squared sine and cosine terms must be equal. Therefore, \[ 200B^2=0.32 \] \[ B^2=0.0016\,\text{m}^2 \] \[ B=0.040\,\text{m}=4\,\text{cm} \] \(\textbf{Constant combined kinetic energy:}\) With this amplitude, \[ K_P+K_Q=0.32[\sin^2(10t)+\cos^2(10t)] \] Using the trigonometric identity, \[ K_P+K_Q=0.32\,\text{J} \] \(\textbf{Potential-energy sum:}\) For each oscillator, \[ U=\frac12m\omega^2x^2 \] Therefore, \[ U_P=\frac12(1)(10)^2(0.08)^2\cos^2(10t) \] \[ U_P=0.32\cos^2(10t)\,\text{J} \] For particle Q, \[ U_Q=\frac12(4)(10)^2(0.04)^2\sin^2(10t) \] \[ U_Q=0.32\sin^2(10t)\,\text{J} \] Adding, \[ U_P+U_Q=0.32[\cos^2(10t)+\sin^2(10t)] \] \[ U_P+U_Q=0.32\,\text{J} \] \(\textbf{Physical interpretation:}\) The two particles execute independent oscillations a quarter-cycle out of phase. When one has maximum kinetic energy, the other has zero kinetic energy. Their individual kinetic energies vary continuously, but the chosen masses and amplitudes make the changes complementary. The same complementary behaviour occurs for their potential energies. Constancy of the combined kinetic energy does not imply that either particle moves at constant speed. \(\textbf{Final answer:}\) \[ \boxed{B=4\,\text{cm},\qquad \sum K=0.32\,\text{J},\qquad \sum U=0.32\,\text{J}} \]

Explanation: \(\textbf{State at the replacement instant:}\) The oscillator is at its positive extreme. Therefore, \[ x_i=+A \] \[ v_i=0 \] The replacement is instantaneous and impulse-free. Hence, the block retains its position and zero velocity immediately after replacement. \(\textbf{Amplitude of the new oscillation:}\) Both springs have equilibrium at \(x=0\). The new oscillator is therefore at displacement \(+A\) with zero velocity. This is an extreme position of its subsequent motion. Thus, \[ A'=A \] \(\textbf{New angular frequency:}\) The original angular frequency is \[ \omega=\sqrt{\frac{k}{m}} \] The replacement force constant is \[ k'=\frac{k}{2} \] Therefore, \[ \omega'=\sqrt{\frac{k'}{m}} \] \[ \omega'=\sqrt{\frac{k}{2m}} \] \[ \omega'=\frac{\omega}{\sqrt2} \] \(\textbf{New period:}\) Since period is inversely proportional to angular frequency, \[ T'=\frac{2\pi}{\omega'} \] \[ T'=\sqrt2\frac{2\pi}{\omega} \] \[ T'=\sqrt2T \] \(\textbf{Initial mechanical energy:}\) The block is at rest at the extreme, so all its mechanical energy is elastic potential energy. \[ E=\frac12kA^2 \] \(\textbf{Mechanical energy immediately after replacement:}\) The block remains at the same displacement, but the new spring has half the stiffness. Therefore, \[ E'=\frac12k'A'^2 \] Substituting, \[ E'=\frac12\left(\frac{k}{2}\right)A^2 \] \[ E'=\frac14kA^2 \] Thus, \[ E'=\frac{E}{2} \] \(\textbf{Work done by the replacement agent:}\) The specified work is the mechanical-energy change during replacement. \[ W_{\mathrm{agent}}=E'-E \] \[ W_{\mathrm{agent}}=\frac{E}{2}-E \] \[ W_{\mathrm{agent}}=-\frac{E}{2} \] \(\textbf{Physical interpretation:}\) At the replacement instant, the block has no kinetic energy. The new spring stores less elastic energy at the same displacement because its force constant is smaller. The external agent therefore removes mechanical energy during the idealised replacement. Afterwards, the new oscillator conserves its own reduced energy. The amplitude remains unchanged because replacement occurs at rest at an extreme, while the period increases because the restoring stiffness decreases. \(\textbf{Final answer:}\) \[ \boxed{A'=A,\qquad T'=\sqrt2T,\qquad W_{\mathrm{agent}}=-\frac{E}{2}} \]

Explanation: \(\textbf{Energy change produced by the impulse:}\) The initial mechanical energy is \[ E=\frac12kA^2 \] The device delivers positive work \[ W=2E \] The spring deformation is unchanged during the instantaneous impulse. Therefore, the supplied work increases the particle's kinetic energy and hence the total spring-block mechanical energy. \[ E'=E+W \] \[ E'=E+2E \] \[ E'=3E \] Statement IV is correct. \(\textbf{Amplitude after the impulse:}\) The spring force constant is unchanged. For the subsequent free SHM, \[ E'=\frac12kA'^2 \] Substituting \(E'=3E\), \[ \frac12kA'^2=3\left(\frac12kA^2\right) \] Cancelling the common factors, \[ A'^2=3A^2 \] Therefore, \[ A'=\sqrt3A \] This result depends on the amount of work added, not on the phase of application. Statement I is correct. \(\textbf{Effect on natural period:}\) For an ideal spring-block oscillator, \[ T=2\pi\sqrt{\frac{m}{k}} \] The impulse does not change \(m\) or \(k\). Hence, the period remains unchanged. Statement II is correct. \(\textbf{Why equal work does not imply equal impulse:}\) Let the signed velocity immediately before an impulse be \(v_i\), and let the signed velocity change be \(\Delta v\). The work of the impulse is \[ W=\frac12m[(v_i+\Delta v)^2-v_i^2] \] Expanding, \[ W=mv_i\Delta v+\frac12m(\Delta v)^2 \] The required \(\Delta v\) depends on the initial signed velocity. Since \(v_i\) varies with oscillation phase, equal positive work does not generally require equal momentum change. Statement III is incorrect. \(\textbf{Physical interpretation:}\) A fixed work input fixes the increase in total mechanical energy, and therefore fixes the new amplitude when the spring constant remains unchanged. A fixed impulse, by contrast, need not deliver the same work at different phases because the block's initial velocity varies. The question distinguishes equal energy addition from equal momentum addition. The correct combination is Statements I, II and IV only.

Explanation: \(\textbf{Connecting restoring force with stored energy:}\) For an ideal spring, \[ F=-kx \] The elastic potential energy is \[ U=\frac12kx^2 \] Taking magnitudes, \[ |F|=k|x| \] Therefore, \[ U=\frac12|F||x| \] \(\textbf{Recovering displacement from simultaneous measurements:}\) The measured force magnitude is \[ |F|=8\,\text{N} \] The elastic potential energy is \[ U=0.16\,\text{J} \] Hence, \[ 0.16=\frac12(8)|x| \] \[ 0.16=4|x| \] \[ |x|=0.04\,\text{m} \] \[ |x|=4\,\text{cm} \] \(\textbf{Determining the force constant:}\) Using Hooke's law, \[ k=\frac{|F|}{|x|} \] Substituting, \[ k=\frac8{0.04} \] \[ k=200\,\text{N m}^{-1} \] The negative force indicates that the measured displacement was positive. \(\textbf{Angular frequency from the measured period:}\) \[ \omega=\frac{2\pi}{T} \] Substituting, \[ \omega=\frac{2\pi}{\pi/5} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Determining the block mass:}\) For spring SHM, \[ \omega^2=\frac{k}{m} \] Therefore, \[ m=\frac{k}{\omega^2} \] Substituting, \[ m=\frac{200}{10^2} \] \[ m=2\,\text{kg} \] \(\textbf{Amplitude in the second trial:}\) The second trial has total mechanical energy \[ E_2=1.00\,\text{J} \] At an extreme, \[ E_2=\frac12kA_2^2 \] Substituting, \[ 1.00=\frac12(200)A_2^2 \] \[ 1.00=100A_2^2 \] \[ A_2^2=0.0100\,\text{m}^2 \] \[ A_2=0.10\,\text{m}=10\,\text{cm} \] \(\textbf{Physical interpretation:}\) Simultaneous force and potential-energy measurements reveal displacement without requiring an initial amplitude measurement. The restoring force gives information proportional to displacement, whereas elastic energy depends on its square. Their combination yields the spring constant. The independently measured period then determines the otherwise unknown mass. Finally, a second trial's total energy determines its amplitude without requiring any information about the phase at which energy was measured. \(\textbf{Final answer:}\) \[ \boxed{|x|=4\,\text{cm},\qquad m=2\,\text{kg},\qquad A_2=10\,\text{cm}} \]

Explanation: \(\textbf{Initial energy distribution:}\) The initial displacement is \[ x_i=+\frac{A}{2} \] For ideal SHM, \[ \frac{U}{E}=\frac{x^2}{A^2} \] Therefore, \[ U_i=\frac{E}{4} \] The initial kinetic energy is \[ K_i=E-U_i \] \[ K_i=\frac{3E}{4} \] \(\textbf{Initial phase from displacement and direction:}\) Write \[ x=A\cos\theta \] Then \[ \cos\theta_i=\frac12 \] The velocity is \[ v=-A\omega\sin\theta \] Because the initial velocity is positive, \[ \sin\theta_i\lt 0 \] We may therefore choose \[ \theta_i=-\frac{\pi}{3} \] \(\textbf{Phase after one-quarter period:}\) During \(T/4\), the phase advances by \[ \Delta\theta=\omega\frac{T}{4} \] \[ \Delta\theta=\frac{\pi}{2} \] Thus, \[ \theta_f=-\frac{\pi}{3}+\frac{\pi}{2} \] \[ \theta_f=\frac{\pi}{6} \] \(\textbf{Final displacement:}\) \[ x_f=A\cos\left(\frac{\pi}{6}\right) \] \[ x_f=+\frac{\sqrt3A}{2} \] \(\textbf{Final potential and kinetic energies:}\) Using the displacement-energy fraction, \[ U_f=E\frac{x_f^2}{A^2} \] \[ U_f=E\frac34 \] \[ U_f=\frac{3E}{4} \] Therefore, \[ K_f=E-U_f \] \[ K_f=\frac{E}{4} \] \(\textbf{Work performed by the spring:}\) By the work-energy theorem, \[ W=K_f-K_i \] Substituting, \[ W=\frac{E}{4}-\frac{3E}{4} \] \[ W=-\frac{E}{2} \] \(\textbf{Kinetic-energy ratio:}\) \[ \frac{K_f}{K_i}=\frac{E/4}{3E/4} \] \[ \frac{K_f}{K_i}=\frac13 \] \(\textbf{Velocity direction at the final instant:}\) At phase \(\pi/6\), \[ v_f=-A\omega\sin\left(\frac{\pi}{6}\right) \] \[ v_f=-\frac{A\omega}{2} \] The velocity is negative. \(\textbf{Physical interpretation:}\) The particle initially moves towards the positive extreme, so the spring opposes its motion and removes kinetic energy. It reaches the positive extreme and reverses before the quarter-period interval ends. At the final instant, it is moving negatively towards equilibrium, but its kinetic energy has not yet recovered to its initial value. The spring's net work over the specified interval is negative because the final displacement magnitude is greater than the initial displacement magnitude. \(\textbf{Final answer:}\) \[ \boxed{W=-\frac{E}{2},\qquad \frac{K_f}{K_i}=\frac13} \] The final velocity is negative.

Explanation: \(\textbf{Connecting restoring force and potential energy:}\) For an ideal spring, \[ F=-kx \] The elastic potential energy is \[ U=\frac12kx^2 \] Squaring the force relation, \[ F^2=k^2x^2 \] From the energy relation, \[ x^2=\frac{2U}{k} \] Substituting, \[ F^2=2kU \] Thus, the slope of an \(F^2\)-versus-\(U\) graph equals \(2k\). \(\textbf{Finding the graph slope:}\) Using the two supplied nonzero points, \[ S=\frac{64-16}{0.32-0.08} \] \[ S=\frac{48}{0.24} \] \[ S=200\,\text{N}^2\text{J}^{-1} \] Therefore, \[ 2k=200 \] \[ k=100\,\text{N m}^{-1} \] \(\textbf{Amplitude from the maximum potential energy:}\) The elastic potential energy is maximum at an extreme position. Therefore, \[ U_{\max}=\frac12kA^2 \] Substituting, \[ 0.32=\frac12(100)A^2 \] \[ 0.32=50A^2 \] \[ A^2=0.0064\,\text{m}^2 \] \[ A=0.08\,\text{m}=8\,\text{cm} \] \(\textbf{Angular frequency from the measured period:}\) \[ \omega=\frac{2\pi}{T} \] Substituting, \[ \omega=\frac{2\pi}{\pi/5} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Determining the block mass:}\) For spring-block SHM, \[ \omega^2=\frac{k}{m} \] Therefore, \[ m=\frac{k}{\omega^2} \] Substituting, \[ m=\frac{100}{10^2} \] \[ m=1\,\text{kg} \] \(\textbf{Graph and unit verification:}\) The maximum restoring-force magnitude should be \[ F_{\max}=kA \] \[ F_{\max}=100(0.08)=8\,\text{N} \] Therefore, \[ F_{\max}^2=64\,\text{N}^2 \] This agrees with the graph endpoint. The slope unit \(\text{N}^2\text{J}^{-1}\) is equivalent to \(\text{N m}^{-1}\), because \(1\,\text{J}=1\,\text{N m}\). \(\textbf{Physical interpretation:}\) Although restoring force changes sign on opposite sides of equilibrium, its square is nonnegative. Consequently, both positive and negative displacements give points on the same straight-line energy graph. The graph slope provides stiffness, the largest potential energy provides amplitude, and the independently measured period provides mass. \(\textbf{Final answer:}\) \[ \boxed{k=100\,\text{N m}^{-1},\qquad A=8\,\text{cm},\qquad m=1\,\text{kg}} \]

Explanation: \(\textbf{Converting the energy condition into displacement:}\) The total mechanical energy is \[ E=K+U \] The condition \(K\gt U\) implies \[ E-U\gt U \] Therefore, \[ U\lt \frac{E}{2} \] For ideal SHM, \[ \frac{U}{E}=\frac{x^2}{A^2} \] Hence, \[ \frac{x^2}{A^2}\lt \frac12 \] \[ |x|\lt \frac{A}{\sqrt2} \] Thus, kinetic energy exceeds potential energy only within a central displacement band. \(\textbf{Fraction of time inside the band:}\) Write \[ x=A\cos\theta \] where \(\theta=\omega t+\phi\). The condition becomes \[ |\cos\theta|\lt \frac1{\sqrt2} \] During the first half-cycle, this holds for \[ \frac{\pi}{4}\lt \theta\lt \frac{3\pi}{4} \] The phase width is \[ \Delta\theta_1=\frac{3\pi}{4}-\frac{\pi}{4} \] \[ \Delta\theta_1=\frac{\pi}{2} \] An identical qualifying interval occurs during the second half-cycle. Therefore, \[ \Delta\theta_{\mathrm{total}}=\pi \] Since the complete cycle corresponds to \(2\pi\), \[ \eta_{\mathrm{time}}=\frac{\pi}{2\pi} \] \[ \eta_{\mathrm{time}}=\frac12 \] \(\textbf{Distance travelled while }K\gt U\textbf{:}\) Let the positive boundary of the permitted region be \[ c=\frac{A}{\sqrt2} \] On one journey from \(+A\) to \(-A\), the particle enters the region at \(+c\) and leaves it at \(-c\). The distance travelled inside the region during this journey is \[ s_1=(+c)-(-c) \] \[ s_1=2c=\sqrt2A \] The return journey from \(-A\) to \(+A\) contributes the same qualifying distance. Therefore, \[ s_{\mathrm{qualifying}}=2\sqrt2A \] \(\textbf{Fraction of total distance:}\) The complete-cycle distance for SHM is \[ s_{\mathrm{cycle}}=4A \] Thus, \[ \eta_{\mathrm{distance}}=\frac{2\sqrt2A}{4A} \] \[ \eta_{\mathrm{distance}}=\frac1{\sqrt2} \] \(\textbf{Physical interpretation:}\) The time fraction and distance fraction need not be equal because the particle does not move with constant speed. The central region is traversed relatively quickly, since the particle's speed is greater nearer equilibrium. Nevertheless, a substantial fraction of the total path lies inside that region. The particle spends half its time but covers more than half its distance where kinetic energy exceeds potential energy. The strict inequality excludes isolated boundary instants, which do not change either fraction. \(\textbf{Final answer:}\) \[ \boxed{\eta_{\mathrm{time}}=\frac12,\qquad \eta_{\mathrm{distance}}=\frac1{\sqrt2}} \]

Explanation: \(\textbf{Common final equilibrium position:}\) Under the final applied force, \[ F_0=kx_0 \] Therefore, \[ x_0=\frac{F_0}{k} \] Substituting, \[ x_0=\frac{10}{100} \] \[ x_0=0.10\,\text{m} \] Both procedures are compared at this same spring extension. \(\textbf{Procedure P: nearly static loading:}\) The applied force increases gradually and approximately balances the spring force at every intermediate displacement. Hence, \[ F_{\mathrm{applied}}\approx kx \] The external work is the area under this increasing force-displacement relation. \[ W_P=\int_0^{x_0}kx\,dx \] \[ W_P=\frac12kx_0^2 \] Substituting, \[ W_P=\frac12(100)(0.10)^2 \] \[ W_P=0.50\,\text{J} \] This energy is stored in the spring. The block has negligible kinetic energy at the end of sufficiently slow loading. \(\textbf{Procedure Q: suddenly applied constant force:}\) The applied force remains \(10\,\text{N}\) throughout the displacement. Therefore, \[ W_Q=F_0x_0 \] Substituting, \[ W_Q=10(0.10) \] \[ W_Q=1.0\,\text{J} \] \(\textbf{Spring energy at the new equilibrium:}\) The spring has the same extension in both procedures. Therefore, \[ U_s=\frac12kx_0^2 \] \[ U_s=0.50\,\text{J} \] \(\textbf{Kinetic energy in Procedure Q:}\) The block starts from rest with zero spring deformation. The work done by the external force becomes the sum of kinetic and elastic potential energies. \[ W_Q=K_Q+U_s \] Hence, \[ K_Q=W_Q-U_s \] \[ K_Q=1.0-0.50 \] \[ K_Q=0.50\,\text{J} \] \(\textbf{Why equilibrium does not imply rest in Procedure Q:}\) At \(x=x_0\), the applied force and spring force balance instantaneously. The net force and acceleration are zero there. However, the block has acquired kinetic energy while moving towards that position. It therefore crosses the new equilibrium with maximum speed rather than stopping. \(\textbf{Physical interpretation:}\) The same final spring extension does not require the same work from the external agent. In quasi-static loading, the force grows with displacement and almost all the supplied work is stored in the spring. In sudden constant-force loading, the applied force does more work over the same displacement, and the excess appears as kinetic energy. The loading history matters even though the spring's final potential energy is identical. \(\textbf{Final answer:}\) \[ \boxed{W_P=0.50\,\text{J},\qquad W_Q=1.0\,\text{J},\qquad K_Q=0.50\,\text{J}} \]

Explanation: \(\textbf{Expressing total energy using velocity and acceleration:}\) For SHM, \[ E=\frac12mv^2+\frac12m\omega^2x^2 \] The signed acceleration satisfies \[ a=-\omega^2x \] Therefore, \[ x=-\frac{a}{\omega^2} \] Substituting into the energy expression, \[ E=\frac12m\left(v^2+\frac{a^2}{\omega^2}\right) \] This expression applies at every instant of the same ideal oscillation. \(\textbf{Equating the energies of states P and Q:}\) Because the motion is undamped, both observations must give the same total energy. \[ v_P^2+\frac{a_P^2}{\omega^2} = v_Q^2+\frac{a_Q^2}{\omega^2} \] Using the numerical readings consistently in centimetres, \[ 40^2+\frac{30^2}{\omega^2} = (-30)^2+\frac{(-40)^2}{\omega^2} \] \[ 1600+\frac{900}{\omega^2} = 900+\frac{1600}{\omega^2} \] Rearranging, \[ 700=\frac{700}{\omega^2} \] Therefore, \[ \omega^2=1\,\text{s}^{-2} \] \[ \omega=1\,\text{rad s}^{-1} \] \(\textbf{Reconstructing the amplitude:}\) At state P, \[ x_P=-\frac{a_P}{\omega^2} \] \[ x_P=-30\,\text{cm} \] Using \[ A^2=x_P^2+\frac{v_P^2}{\omega^2} \] we obtain \[ A^2=30^2+40^2 \] \[ A^2=2500\,\text{cm}^2 \] \[ A=50\,\text{cm} \] \(\textbf{Total mechanical energy:}\) Convert the amplitude to metres. \[ A=0.50\,\text{m} \] The total energy is \[ E=\frac12m\omega^2A^2 \] Substituting, \[ E=\frac12(1)(1)^2(0.50)^2 \] \[ E=0.125\,\text{J} \] \(\textbf{Position and direction at state P:}\) The acceleration at state P is positive. Since acceleration points towards equilibrium, the displacement must be negative. The measured velocity is also positive, so the particle is moving towards equilibrium with increasing speed. \(\textbf{Independent consistency check at state Q:}\) At Q, \[ x_Q=-\frac{-40}{1} \] \[ x_Q=+40\,\text{cm} \] The amplitude relation gives \[ A^2=40^2+(-30)^2 \] \[ A=50\,\text{cm} \] Thus, both observations describe the same oscillator with the same amplitude and energy. \(\textbf{Physical interpretation:}\) A single simultaneous velocity and acceleration reading generally cannot determine both amplitude and angular frequency. Two energetically linked states provide an additional equation that eliminates the unknown amplitude. The signs of acceleration then determine displacement signs independently of the energy calculation. \(\textbf{Final answer:}\) \[ \boxed{\omega=1\,\text{rad s}^{-1},\qquad E=0.125\,\text{J}} \] The displacement at P is negative.