Oscillations and Waves MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Interpreting the graph intercepts:}\) The horizontal intercepts correspond to the maximum positive and negative velocities. Therefore, \[ v_{\max}=12\,\text{cm s}^{-1} \] The vertical intercepts correspond to the maximum positive and negative accelerations. Hence, \[ a_{\max}=36\,\text{cm s}^{-2} \] \(\textbf{Connecting maximum speed and acceleration:}\) For SHM, \[ v_{\max}=A\omega \] and \[ a_{\max}=A\omega^2 \] Dividing the second relation by the first, \[ \frac{a_{\max}}{v_{\max}}=\omega \] Therefore, \[ \omega=\frac{36}{12} \] \[ \omega=3\,\text{rad s}^{-1} \] \(\textbf{Determining the amplitude:}\) Using \[ v_{\max}=A\omega \] we obtain \[ A=\frac{v_{\max}}{\omega} \] Substituting, \[ A=\frac{12}{3} \] \[ A=4\,\text{cm} \] \(\textbf{Finding the acceleration magnitude at the specified speed:}\) The SHM speed relation is \[ v^2=\omega^2(A^2-x^2) \] Given, \[ v=+6\,\text{cm s}^{-1} \] Therefore, \[ 6^2=3^2(4^2-x^2) \] \[ 36=9(16-x^2) \] Dividing by \(9\), \[ 4=16-x^2 \] \[ x^2=12\,\text{cm}^2 \] Hence, \[ |x|=2\sqrt3\,\text{cm} \] \(\textbf{Selecting the displacement sign:}\) The particle is specified to have positive displacement. Therefore, \[ x=+2\sqrt3\,\text{cm} \] \(\textbf{Calculating signed acceleration:}\) For SHM, \[ a=-\omega^2x \] Substituting, \[ a=-(3)^2(2\sqrt3) \] \[ a=-18\sqrt3\,\text{cm s}^{-2} \] \(\textbf{Physical interpretation:}\) The acceleration is negative because the particle lies on the positive side of equilibrium. Its velocity is positive, so it is moving away from equilibrium towards the positive extreme. Therefore, the restoring acceleration opposes its velocity and reduces its speed. The ellipse has opposite acceleration values for the two possible displacement signs at a given speed magnitude. The extra information about displacement sign is essential for selecting the correct acceleration branch. \(\textbf{Final answer:}\) \[ \boxed{\omega=3\,\text{rad s}^{-1},\qquad A=4\,\text{cm},\qquad a=-18\sqrt3\,\text{cm s}^{-2}} \]

Explanation: \(\textbf{Original restoring-force law:}\) For an ideal spring of force constant \(k\), the restoring force measured from its equilibrium position is \[ F=-kx \] A force-versus-displacement graph is therefore a straight line through the origin with negative slope \(-k\). \(\textbf{Effect of the applied constant force:}\) The additional force is independent of position. The new net force is \[ F_{\mathrm{net}}=F_0-kx \] This remains a linear function of \(x\). Its slope is still \(-k\), so the original and new force graphs are parallel. Therefore, Statement I is correct. \(\textbf{Evaluating the slope claim:}\) The constant force changes the vertical intercept but does not alter the coefficient of \(x\). The slope is neither more negative nor less negative. Statement II is incorrect. \(\textbf{Determining the new zero-force position:}\) At the new equilibrium, \[ F_{\mathrm{net}}=0 \] Therefore, \[ F_0-kx_0=0 \] \[ x_0=\frac{F_0}{k} \] Since \(F_0\) is positive and \(k\gt 0\), the new equilibrium lies on the positive side of the original equilibrium. Statement III is correct. \(\textbf{Testing the natural-frequency claim:}\) Measure displacement \(y\) from the new equilibrium. Then \[ x=x_0+y \] Substituting into the net-force equation, \[ F_{\mathrm{net}}=F_0-k(x_0+y) \] Since \(F_0=kx_0\), \[ F_{\mathrm{net}}=-ky \] Newton's second law gives \[ ma=-ky \] Therefore, \[ a=-\frac{k}{m}y \] The angular frequency remains \[ \omega=\sqrt{\frac{k}{m}} \] Statement IV is incorrect. \(\textbf{Physical interpretation:}\) A constant external force shifts the equilibrium position without changing the stiffness of an ideal spring. On the graph, this appears as a parallel translation of the force line. The new zero-force coordinate changes, but the negative slope remains the same. The correct combination is Statements I and III only.

Explanation: \(\textbf{Angular frequency of the block:}\) For a spring-block oscillator, \[ \omega=\sqrt{\frac{k}{m}} \] Substituting, \[ \omega=\sqrt{\frac{200}{2}} \] \[ \omega=\sqrt{100} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Amplitude from the force maximum:}\) The restoring-force magnitude is maximum at an extreme position. Therefore, \[ F_{\max}=kA \] Given, \[ F_{\max}=8\,\text{N} \] Hence, \[ A=\frac{F_{\max}}{k} \] \[ A=\frac{8}{200} \] \[ A=0.04\,\text{m} \] \(\textbf{Maximum speed:}\) For SHM, \[ v_{\max}=A\omega \] Substituting, \[ v_{\max}=0.04(10) \] \[ v_{\max}=0.40\,\text{m s}^{-1} \] \(\textbf{Position when restoring force vanishes:}\) The restoring force satisfies \[ F=-kx \] At \(t=0\), the force is zero. Therefore, \[ x=0 \] The block is at equilibrium, where its speed magnitude is maximum. \(\textbf{Inferring the velocity sign from the graph slope:}\) Differentiate the restoring-force equation with respect to time. \[ \frac{dF}{dt}=-k\frac{dx}{dt} \] Therefore, \[ \frac{dF}{dt}=-kv \] The force-time graph has positive slope at \(t=0\). Thus, \[ \frac{dF}{dt}\gt 0 \] Since \(k\gt 0\), \[ v\lt 0 \] Hence, \[ v(0)=-0.40\,\text{m s}^{-1} \] \(\textbf{Time to the positive force maximum:}\) Immediately after \(t=0\), the block moves into negative displacement. The restoring force becomes positive and increases in magnitude. The positive force maximum occurs when the block reaches the negative extreme. The journey from equilibrium to an extreme takes one-quarter period. \[ \Delta t=\frac{T}{4} \] Using \[ T=\frac{2\pi}{\omega} \] we obtain \[ \Delta t=\frac{\pi}{2\omega} \] Substituting \(\omega=10\), \[ \Delta t=\frac{\pi}{20}\,\text{s} \] \(\textbf{Physical interpretation:}\) A force-time graph's positive slope does not mean the particle has positive velocity. For a restoring force, force and displacement have opposite signs. Consequently, the rate of change of force has the opposite sign to velocity. The positive force maximum occurs at the negative displacement extreme, not at the positive extreme. \(\textbf{Final answer:}\) \[ \boxed{v(0)=-0.40\,\text{m s}^{-1},\qquad \Delta t=\frac{\pi}{20}\,\text{s}} \]

Explanation: \(\textbf{Mass relationship:}\) The given masses satisfy \[ m_Q=4m_P \] The two oscillators have equal periods. Therefore, their angular frequencies are also equal. \[ \omega_Q=\omega_P \] \(\textbf{Relating stiffness and mass:}\) For a spring-block oscillator, \[ \omega^2=\frac{k}{m} \] Equality of angular frequencies implies \[ \frac{k_Q}{m_Q}=\frac{k_P}{m_P} \] Rearranging, \[ \frac{k_Q}{k_P}=\frac{m_Q}{m_P} \] Substituting the mass ratio, \[ \frac{k_Q}{k_P}=4 \] Therefore, \[ k_Q:k_P=4:1 \] \(\textbf{Using equality of maximum restoring forces:}\) The maximum restoring-force magnitude is \[ F_{\max}=kA \] The two maximum forces are equal. Hence, \[ k_QA_Q=k_PA_P \] Therefore, \[ \frac{A_Q}{A_P}=\frac{k_P}{k_Q} \] Substituting, \[ \frac{A_Q}{A_P}=\frac14 \] Thus, \[ A_Q:A_P=1:4 \] \(\textbf{Comparing maximum speeds:}\) The maximum speed is \[ v_{\max}=A\omega \] Since both oscillators have the same angular frequency, \[ \frac{v_{\max,Q}}{v_{\max,P}} =\frac{A_Q}{A_P} \] Therefore, \[ \frac{v_{\max,Q}}{v_{\max,P}}=\frac14 \] \[ v_{\max,Q}:v_{\max,P}=1:4 \] \(\textbf{Physical interpretation:}\) The heavier block requires a proportionally stiffer spring to preserve the same natural period. Because the maximum restoring forces are also constrained to be equal, the stiffer oscillator must have a smaller amplitude. Equal periods do not require equal amplitudes or maximum speeds. The simultaneous constraints determine all three ratios without requiring the actual values of the masses or spring constants. \(\textbf{Final answer:}\) \[ \boxed{k_Q:k_P=4:1,\qquad A_Q:A_P=1:4,\qquad v_{\max,Q}:v_{\max,P}=1:4} \]

Explanation: \(\textbf{Static equilibrium condition:}\) For a mass \(m\) hanging from an ideal vertical spring, \[ kl=mg \] where \(l\) is the static equilibrium extension. Therefore, \[ l=\frac{mg}{k} \] \(\textbf{Comparing extensions at the two locations:}\) The block mass and spring constant remain unchanged. Hence, \[ l\propto g \] Therefore, \[ \frac{l_Q}{l_P}=\frac{g_Q}{g_P} \] Substituting, \[ \frac{l_Q}{10\,\text{cm}}=\frac{5}{10} \] Thus, \[ l_Q=5\,\text{cm} \] \(\textbf{Time period of vertical spring oscillations:}\) For a vertical spring-block oscillator, displacement measured from the equilibrium position satisfies \[ ma=-kx \] Therefore, \[ \omega=\sqrt{\frac{k}{m}} \] The time period is \[ T=2\pi\sqrt{\frac{m}{k}} \] \(\textbf{Using the measured static extension:}\) From the equilibrium condition, \[ kl_P=mg_P \] Rearranging, \[ \frac{m}{k}=\frac{l_P}{g_P} \] Therefore, \[ T=2\pi\sqrt{\frac{l_P}{g_P}} \] \(\textbf{Calculating the period at }P\textbf{:}\) The measured extension is \[ l_P=0.10\,\text{m} \] Therefore, \[ T_P=2\pi\sqrt{\frac{0.10}{10}} \] \[ T_P=2\pi\sqrt{0.01} \] \[ T_P=2\pi(0.10) \] \[ T_P=\frac{\pi}{5}\,\text{s} \] \(\textbf{Period at the new location:}\) At location \(Q\), neither \(m\) nor \(k\) changes. Consequently, \[ T_Q=T_P \] \[ T_Q=\frac{\pi}{5}\,\text{s} \] \(\textbf{Independent verification using the new extension:}\) \[ T_Q=2\pi\sqrt{\frac{l_Q}{g_Q}} \] Substituting, \[ T_Q=2\pi\sqrt{\frac{0.05}{5}} \] \[ T_Q=2\pi\sqrt{0.01} \] \[ T_Q=\frac{\pi}{5}\,\text{s} \] \(\textbf{Physical interpretation:}\) Reducing gravitational acceleration reduces the static spring extension. However, the natural period remains unchanged because it depends only on the block mass and spring stiffness. The relation \(T=2\pi\sqrt{l/g}\) is valid when \(l\) is the static extension produced by the same local gravitational acceleration \(g\). The extension and gravitational acceleration change in the same proportion, leaving their ratio unchanged. \(\textbf{Final answer:}\) \[ \boxed{l_Q=5\,\text{cm},\qquad T_Q=\frac{\pi}{5}\,\text{s}} \]

Explanation: \(\textbf{Spring forces at the reference coordinate:}\) At \(X=0\), the left spring is stretched by \(0.10\,\text{m}\). It pulls the block towards the left. Therefore, \[ F_L(0)=-k_Ll_L \] \[ F_L(0)=-60(0.10) \] \[ F_L(0)=-6\,\text{N} \] The right spring is stretched by \(0.05\,\text{m}\). It pulls the block towards the right. Therefore, \[ F_R(0)=+k_Rl_R \] \[ F_R(0)=40(0.05) \] \[ F_R(0)=+2\,\text{N} \] \(\textbf{Net force at }X=0\textbf{:}\) \[ F_{\mathrm{net}}(0)=F_L(0)+F_R(0) \] \[ F_{\mathrm{net}}(0)=-6+2 \] \[ F_{\mathrm{net}}(0)=-4\,\text{N} \] Thus, the block initially accelerates towards the left. \(\textbf{Force law at a general coordinate:}\) When the block moves a distance \(X\) towards the right, the left spring's extension increases by \(X\). Therefore, \[ F_L=-60(0.10+X) \] The right spring's extension decreases by \(X\). Hence, \[ F_R=40(0.05-X) \] The net force is \[ F_{\mathrm{net}}=-60(0.10+X)+40(0.05-X) \] Expanding, \[ F_{\mathrm{net}}=-6-60X+2-40X \] \[ F_{\mathrm{net}}=-4-100X \] \(\textbf{Finding the equilibrium coordinate:}\) At equilibrium, \[ F_{\mathrm{net}}=0 \] Therefore, \[ -4-100X_0=0 \] \[ X_0=-0.04\,\text{m} \] \[ X_0=-4\,\text{cm} \] \(\textbf{Restoring-force coefficient:}\) Let \[ x=X-X_0 \] The net force can be written as \[ F_{\mathrm{net}}=-100x \] Therefore, the effective spring constant is \[ k_{\mathrm{eff}}=100\,\text{N m}^{-1} \] \(\textbf{Angular frequency:}\) \[ \omega=\sqrt{\frac{k_{\mathrm{eff}}}{m}} \] \[ \omega=\sqrt{\frac{100}{1}} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Amplitude from the initial condition:}\) The block is released from rest at \(X=0\). Its displacement from equilibrium is \[ x_i=0-(-0.04) \] \[ x_i=+0.04\,\text{m} \] Since its initial velocity is zero, it begins at an extreme position. Therefore, \[ A=0.04\,\text{m}=4\,\text{cm} \] \(\textbf{Maximum speed:}\) \[ v_{\max}=A\omega \] \[ v_{\max}=0.04(10) \] \[ v_{\max}=0.40\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) The springs are already stretched at the chosen coordinate origin. Their forces at \(X=0\) are unequal, so this coordinate is not the equilibrium position. The initial net force points left, and the new equilibrium lies to the left of the chosen origin. The effective stiffness is the sum of the two spring constants because both springs contribute restoring force for displacement away from equilibrium. \(\textbf{Final answer:}\) \[ \boxed{X_0=-4\,\text{cm},\qquad A=4\,\text{cm},\qquad v_{\max}=0.40\,\text{m s}^{-1}} \]

Explanation: \(\textbf{Original oscillator:}\) The original angular frequency is \[ \omega=\sqrt{\frac{k}{m}} \] The block executes SHM with amplitude \(A\). At the attachment instant, \[ x=\frac{A}{\sqrt2} \] \(\textbf{Initial velocity at the attachment instant:}\) The SHM speed relation is \[ v^2=\omega^2(A^2-x^2) \] Substituting, \[ v^2=\omega^2\left(A^2-\frac{A^2}{2}\right) \] \[ v^2=\frac{A^2\omega^2}{2} \] Therefore, \[ |v|=\frac{A\omega}{\sqrt2} \] Since the velocity is positive, \[ v=+\frac{A\omega}{\sqrt2} \] \(\textbf{Mass after attachment:}\) The added mass is \(3m\). Therefore, the new oscillating mass is \[ m'=m+3m \] \[ m'=4m \] \(\textbf{New angular frequency:}\) The spring force constant remains unchanged. Therefore, \[ \omega'=\sqrt{\frac{k}{m'}} \] Substituting, \[ \omega'=\sqrt{\frac{k}{4m}} \] \[ \omega'=\frac12\sqrt{\frac{k}{m}} \] \[ \omega'=\frac{\omega}{2} \] \(\textbf{New period:}\) \[ T'=\frac{2\pi}{\omega'} \] \[ T'=\frac{2\pi}{\omega/2} \] \[ T'=2T \] \(\textbf{Calculating the new amplitude:}\) The displacement and velocity immediately after attachment remain unchanged. For the new SHM, \[ v^2=\omega'^2(A'^2-x^2) \] Rearranging, \[ A'^2=x^2+\frac{v^2}{\omega'^2} \] Substituting the known state, \[ A'^2=\frac{A^2}{2} +\frac{A^2\omega^2/2}{\omega^2/4} \] Simplifying the second term, \[ A'^2=\frac{A^2}{2}+2A^2 \] \[ A'^2=\frac{5A^2}{2} \] Therefore, \[ A'=A\sqrt{\frac52} \] \(\textbf{Physical interpretation:}\) Adding the co-moving mass increases inertia without changing the spring stiffness. Consequently, the natural angular frequency decreases and the period increases. The instantaneous speed is unchanged at attachment, but the new lower angular frequency requires a larger amplitude to accommodate the same position and velocity. The new amplitude exceeds the original amplitude even though no instantaneous displacement occurs during attachment. \(\textbf{Final answer:}\) \[ \boxed{A'=A\sqrt{\frac52},\qquad T'=2T} \]

Explanation: \(\textbf{Original oscillator parameters:}\) The original block mass is \[ m_1=2\,\text{kg} \] The original spring constant is \[ k_1=50\,\text{N m}^{-1} \] Its angular frequency is \[ \omega_1=\sqrt{\frac{k_1}{m_1}} \] Substituting, \[ \omega_1=\sqrt{\frac{50}{2}} \] \[ \omega_1=\sqrt{25} \] \[ \omega_1=5\,\text{rad s}^{-1} \] \(\textbf{Modified oscillator mass:}\) An additional mass of \(3\,\text{kg}\) is attached. Therefore, \[ m_2=2+3 \] \[ m_2=5\,\text{kg} \] \(\textbf{Required effective stiffness:}\) The modified oscillator must have the same period. Therefore, its angular frequency must remain \(5\,\text{rad s}^{-1}\). For the modified oscillator, \[ \omega_2^2=\frac{k_{\mathrm{eff}}}{m_2} \] Rearranging, \[ k_{\mathrm{eff}}=m_2\omega_2^2 \] Substituting, \[ k_{\mathrm{eff}}=5(5)^2 \] \[ k_{\mathrm{eff}}=125\,\text{N m}^{-1} \] \(\textbf{Force constant of the additional spring:}\) Springs connected in parallel have additive force constants. Therefore, \[ k_{\mathrm{eff}}=k_1+k_{\mathrm{add}} \] Hence, \[ k_{\mathrm{add}}=k_{\mathrm{eff}}-k_1 \] \[ k_{\mathrm{add}}=125-50 \] \[ k_{\mathrm{add}}=75\,\text{N m}^{-1} \] \(\textbf{Original maximum restoring force:}\) The original oscillator has amplitude \(A\). Therefore, \[ F_{\max,1}=k_1A \] \[ F_{\max,1}=50A \] \(\textbf{Modified maximum restoring force:}\) The modified oscillator is also released from rest at displacement \(A\). Therefore, its amplitude is \(A\). Its maximum restoring force is \[ F_{\max,2}=k_{\mathrm{eff}}A \] \[ F_{\max,2}=125A \] \(\textbf{Force comparison:}\) \[ \frac{F_{\max,2}}{F_{\max,1}} =\frac{125A}{50A} \] \[ \frac{F_{\max,2}}{F_{\max,1}}=\frac52 \] \(\textbf{Physical interpretation:}\) Adding mass alone would increase the oscillator's time period. To retain the original period, the effective spring stiffness must increase in the same proportion as the total mass. The additional parallel spring provides the required stiffness. Although the periods and amplitudes are equal, the maximum restoring force becomes greater because the modified spring combination is stiffer. \(\textbf{Final answer:}\) \[ \boxed{k_{\mathrm{add}}=75\,\text{N m}^{-1},\qquad F_{\max,2}:F_{\max,1}=5:2} \]

Explanation: \(\textbf{Using Newton's second law:}\) The restoring force is the net horizontal force on the block. Therefore, \[ F=ma \] The block mass can be determined directly from simultaneous force and acceleration measurements. \(\textbf{Mass from the first observation:}\) Given, \[ F=-8\,\text{N} \] \[ a=-16\,\text{m s}^{-2} \] Therefore, \[ m=\frac{F}{a} \] \[ m=\frac{-8}{-16} \] \[ m=0.50\,\text{kg} \] \(\textbf{Verifying the mass with the second observation:}\) For the second measurement, \[ m=\frac{+4}{+8} \] \[ m=0.50\,\text{kg} \] Both observations give the same mass, confirming their consistency with Newton's second law. \(\textbf{Determining the spring force constant:}\) Hooke's law gives \[ F=-kx \] Using the first observation, \[ -8=-k(0.04) \] Therefore, \[ k=\frac{8}{0.04} \] \[ k=200\,\text{N m}^{-1} \] \(\textbf{Checking the second force reading:}\) At \[ x=-0.02\,\text{m} \] the predicted restoring force is \[ F=-200(-0.02) \] \[ F=+4\,\text{N} \] This agrees with the second observation. \(\textbf{Calculating angular frequency:}\) For a spring-block oscillator, \[ \omega=\sqrt{\frac{k}{m}} \] Substituting, \[ \omega=\sqrt{\frac{200}{0.50}} \] \[ \omega=\sqrt{400} \] \[ \omega=20\,\text{rad s}^{-1} \] \(\textbf{Time period:}\) \[ T=\frac{2\pi}{\omega} \] \[ T=\frac{2\pi}{20} \] \[ T=\frac{\pi}{10}\,\text{s} \] \(\textbf{Physical interpretation:}\) The force-to-acceleration ratio gives the mass because force and acceleration are related by Newton's second law. The restoring-force-to-displacement ratio gives the spring stiffness after accounting for the negative sign in Hooke's law. These two independent relationships allow both unknown mechanical parameters to be determined. The period then follows from the ratio of mass to stiffness. The negative restoring forces at positive displacements and positive forces at negative displacements are consistent with stable equilibrium. \(\textbf{Final answer:}\) \[ \boxed{m=0.50\,\text{kg},\qquad k=200\,\text{N m}^{-1},\qquad T=\frac{\pi}{10}\,\text{s}} \]

Explanation: \(\textbf{Initial state:}\) The block starts at the positive extreme. Therefore, \[ x_i=+A \] \[ v_i=0 \] The amplitude and angular frequency are \[ A=0.10\,\text{m} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Velocity at equilibrium:}\) The particle moves from the positive extreme towards equilibrium. Its velocity is therefore negative. At equilibrium, its speed magnitude is maximum. \[ v_{\max}=A\omega \] Substituting, \[ v_{\max}=0.10(10) \] \[ v_{\max}=1.0\,\text{m s}^{-1} \] Thus, its signed velocity at equilibrium is \[ v_f=-1.0\,\text{m s}^{-1} \] \(\textbf{Impulse from momentum change:}\) The impulse of the net force is equal to the change in momentum. \[ J=\Delta p \] \[ J=m(v_f-v_i) \] Substituting, \[ J=0.20(-1.0-0) \] \[ J=-0.20\,\text{N s} \] The negative sign indicates that the net impulse acts towards the negative \(x\)-direction. \(\textbf{Time taken to reach equilibrium:}\) The SHM time period is \[ T=\frac{2\pi}{\omega} \] \[ T=\frac{2\pi}{10} \] \[ T=\frac{\pi}{5}\,\text{s} \] The journey from an extreme to equilibrium takes one-quarter period. \[ \Delta t=\frac{T}{4} \] \[ \Delta t=\frac{\pi}{20}\,\text{s} \] \(\textbf{Average restoring force:}\) Average force over the interval is the total impulse divided by elapsed time. \[ F_{\mathrm{avg}}=\frac{J}{\Delta t} \] Substituting, \[ F_{\mathrm{avg}}=\frac{-0.20}{\pi/20} \] \[ F_{\mathrm{avg}}=-\frac4\pi\,\text{N} \] \(\textbf{Initial restoring force:}\) The effective force constant follows from \[ k=m\omega^2 \] Substituting, \[ k=0.20(10)^2 \] \[ k=20\,\text{N m}^{-1} \] At the initial positive extreme, \[ F_i=-kA \] \[ F_i=-20(0.10) \] \[ F_i=-2\,\text{N} \] \(\textbf{Comparing force magnitudes:}\) \[ \frac{|F_{\mathrm{avg}}|}{|F_i|} =\frac{4/\pi}{2} \] \[ \frac{|F_{\mathrm{avg}}|}{|F_i|}=\frac2\pi \] \(\textbf{Physical interpretation:}\) The restoring force is maximum in magnitude at the initial extreme. Its magnitude decreases continuously as the block approaches equilibrium. At equilibrium, the instantaneous restoring force becomes zero, even though the block reaches maximum speed. The average restoring-force magnitude during the journey is therefore smaller than its initial magnitude. Impulse depends on momentum change, so it can be calculated without directly integrating the varying force. \(\textbf{Final answer:}\) \[ \boxed{J=-0.20\,\text{N s},\qquad \frac{|F_{\mathrm{avg}}|}{|F_i|}=\frac2\pi} \]

Explanation: \(\textbf{Different restoring laws on the two sides:}\) The restoring force is proportional to displacement on each side of equilibrium, but the proportionality coefficients are different. For positive displacement, \[ k_+=100\,\text{N m}^{-1} \] For negative displacement, \[ k_-=400\,\text{N m}^{-1} \] The motion is therefore sinusoidal within each half-region, but it is not one SHM described by a single angular frequency throughout the complete cycle. \(\textbf{Angular frequency in the positive region:}\) \[ \omega_+=\sqrt{\frac{k_+}{m}} \] \[ \omega_+=\sqrt{\frac{100}{1}} \] \[ \omega_+=10\,\text{rad s}^{-1} \] \(\textbf{Speed on reaching equilibrium:}\) The particle starts from rest at \[ A_+=0.08\,\text{m} \] The maximum speed during its positive-side motion occurs at equilibrium. \[ v_0=\omega_+A_+ \] \[ v_0=10(0.08) \] \[ v_0=0.80\,\text{m s}^{-1} \] \(\textbf{Motion in the negative region:}\) The angular frequency associated with the negative-side restoring law is \[ \omega_-=\sqrt{\frac{k_-}{m}} \] \[ \omega_-=\sqrt{\frac{400}{1}} \] \[ \omega_-=20\,\text{rad s}^{-1} \] The particle crosses equilibrium with continuous velocity, so its speed on entering the negative region is still \(0.80\,\text{m s}^{-1}\). For the negative-side SHM segment, \[ v_0=\omega_-A_- \] Therefore, \[ A_-=\frac{v_0}{\omega_-} \] \[ A_-=\frac{0.80}{20} \] \[ A_-=0.04\,\text{m}=4\,\text{cm} \] \(\textbf{Time spent in the positive region:}\) A journey from the positive extreme to equilibrium takes one-quarter of the period associated with \(\omega_+\). The return journey from equilibrium to the positive extreme takes the same time. Thus, \[ t_+=2\left(\frac{\pi}{2\omega_+}\right) \] \[ t_+=\frac{\pi}{10}\,\text{s} \] \(\textbf{Time spent in the negative region:}\) Similarly, the motion from equilibrium to the negative extreme and back takes \[ t_-=\frac{\pi}{\omega_-} \] \[ t_-=\frac{\pi}{20}\,\text{s} \] \(\textbf{Complete time period:}\) \[ T=t_++t_- \] \[ T=\frac{\pi}{10}+\frac{\pi}{20} \] \[ T=\frac{3\pi}{20}\,\text{s} \] \(\textbf{Physical interpretation:}\) The stronger restoring force on the negative side produces a smaller negative excursion and a shorter time spent in that region. The particle returns to its original position and velocity after the calculated interval. Its motion is periodic, but the unequal restoring-force coefficients prevent it from being a single ideal SHM across both sides of equilibrium. \(\textbf{Final answer:}\) \[ \boxed{A_-=4\,\text{cm},\qquad T=\frac{3\pi}{20}\,\text{s}} \]

Explanation: \(\textbf{Phase information from the common displacement:}\) At \(t=0\), \[ x_P=x_Q=\frac{A}{2} \] For either particle, \[ A\cos\phi=\frac{A}{2} \] Therefore, \[ \cos\phi=\frac12 \] The two possible phase values within one cycle are \[ \phi=\frac{\pi}{3},\qquad \phi=\frac{5\pi}{3} \] \(\textbf{Selecting the phase of particle }P\textbf{:}\) For cosine-form SHM, \[ v=-A\omega\sin(\omega t+\phi) \] Particle \(P\) initially has positive velocity. Therefore, \[ \sin\phi_P\lt 0 \] This selects \[ \phi_P=\frac{5\pi}{3} \] Equivalently, its phase constant can be written as \(-\pi/3\). \(\textbf{Selecting the phase of particle }Q\textbf{:}\) Particle \(Q\) initially has negative velocity. Therefore, \[ \sin\phi_Q\gt 0 \] Hence, \[ \phi_Q=\frac{\pi}{3} \] \(\textbf{Smallest positive phase separation:}\) The directed difference using these phase constants is \[ \phi_Q-\phi_P=\frac{\pi}{3}-\frac{5\pi}{3} \] \[ \phi_Q-\phi_P=-\frac{4\pi}{3} \] Phases differing by integral multiples of \(2\pi\) represent the same oscillatory state. Adding \(2\pi\), \[ \Delta\phi=-\frac{4\pi}{3}+2\pi \] \[ \Delta\phi=\frac{2\pi}{3} \] Thus, the smallest positive phase difference is \(2\pi/3\). \(\textbf{Phase advance after }T/6\textbf{:}\) During the interval, \[ \Delta\theta=\omega\Delta t \] \[ \Delta\theta=\frac{2\pi}{T}\frac{T}{6} \] \[ \Delta\theta=\frac{\pi}{3} \] \(\textbf{Final position of }P\textbf{:}\) \[ x_P=A\cos\left(\frac{5\pi}{3}+\frac{\pi}{3}\right) \] \[ x_P=A\cos(2\pi) \] \[ x_P=+A \] \(\textbf{Final position of }Q\textbf{:}\) \[ x_Q=A\cos\left(\frac{\pi}{3}+\frac{\pi}{3}\right) \] \[ x_Q=A\cos\left(\frac{2\pi}{3}\right) \] \[ x_Q=-\frac{A}{2} \] \(\textbf{Physical interpretation:}\) Equal initial displacement does not imply equal phase. The opposite initial velocity directions select different branches of the same cosine value. After time \(T/6\), one particle reaches its positive extreme while the other has crossed equilibrium into the negative half of its path. The phase difference remains constant because their angular frequencies are identical. \(\textbf{Final answer:}\) \[ \boxed{\Delta\phi=\frac{2\pi}{3},\qquad x_P(T/6)=+A,\quad x_Q(T/6)=-\frac{A}{2}} \]

Explanation: \(\textbf{Static equilibrium extension:}\) At equilibrium, the upward force of the spring balances the block's weight. \[ kl_0=mg \] Therefore, \[ l_0=\frac{mg}{k} \] Substituting, \[ l_0=\frac{0.50(10)}{100} \] \[ l_0=0.05\,\text{m}=5\,\text{cm} \] \(\textbf{Spring extension at the upper extreme:}\) The block oscillates with amplitude \[ A=2\,\text{cm} \] At the upper extreme, it is \(2\,\text{cm}\) above equilibrium. Therefore, its spring extension is \[ l_u=l_0-A \] \[ l_u=5-2=3\,\text{cm} \] \(\textbf{Spring force at the upper extreme:}\) \[ F_{s,u}=kl_u \] \[ F_{s,u}=100(0.03) \] \[ F_{s,u}=3\,\text{N} \] The spring force acts upward. The block's weight is \[ mg=0.50(10)=5\,\text{N} \] The weight exceeds the upward spring force. Thus, the net force is directed downward, with magnitude \[ F_{\mathrm{net},u}=5-3=2\,\text{N} \] \(\textbf{Spring extension at the lower extreme:}\) At the lower extreme, the block is \(2\,\text{cm}\) below equilibrium. Therefore, \[ l_l=l_0+A \] \[ l_l=5+2=7\,\text{cm} \] \(\textbf{Spring force at the lower extreme:}\) \[ F_{s,l}=kl_l \] \[ F_{s,l}=100(0.07) \] \[ F_{s,l}=7\,\text{N} \] The upward spring force now exceeds the weight. Therefore, the net force acts upward, with magnitude \[ F_{\mathrm{net},l}=7-5=2\,\text{N} \] \(\textbf{Distinguishing spring force from restoring force:}\) The spring force acts upward at both extreme positions because the spring remains extended throughout the motion. However, the net restoring force is directed towards equilibrium and therefore reverses direction between the extremes. It is the net force, not the spring force alone, that obeys the SHM relation \(F_{\mathrm{net}}=-kx\) when \(x\) is measured from the equilibrium position. \(\textbf{Physical check:}\) The net restoring-force magnitudes at the two extremes are both \(kA=2\,\text{N}\). Their directions are opposite, consistent with equal-amplitude oscillation about equilibrium. \(\textbf{Final answer:}\) At the upper extreme, the spring force is \(3\,\text{N}\) upward and the net force is downward. At the lower extreme, the spring force is \(7\,\text{N}\) upward and the net force is upward.

Explanation: \(\textbf{Initial phase and direction:}\) Represent the displacement by \[ x=A\cos\theta \] At the initial instant, \[ x_0=\frac{A}{2} \] Therefore, \[ \cos\theta_0=\frac12 \] The particle is moving towards the positive extreme, so its velocity is positive. Since \[ v=-A\omega\sin\theta \] the sine of the initial phase must be negative. Hence, \[ \theta_0=\frac{5\pi}{3} \] \(\textbf{Phase advance during }T/6\textbf{:}\) \[ \Delta\theta=\omega\Delta t \] \[ \Delta\theta=\frac{2\pi}{T}\frac{T}{6} \] \[ \Delta\theta=\frac{\pi}{3} \] \(\textbf{Final phase:}\) \[ \theta_f=\theta_0+\Delta\theta \] \[ \theta_f=\frac{5\pi}{3}+\frac{\pi}{3} \] \[ \theta_f=2\pi \] \(\textbf{Final displacement:}\) \[ x_f=A\cos(2\pi) \] \[ x_f=+A \] \(\textbf{Final velocity and speed:}\) \[ v_f=-A\omega\sin(2\pi) \] \[ v_f=0 \] The particle is momentarily at rest at the positive extreme. Before reaching this extreme, the particle has positive velocity while its restoring acceleration is negative, so its speed decreases continuously towards zero. \(\textbf{Interpreting the instant precisely:}\) At the specified final instant, the instantaneous rate of change of speed requires care because speed reaches its minimum zero value at a turning point. Immediately before the instant, speed is decreasing; immediately after, speed begins increasing in the opposite direction. Thus, describing speed simply as increasing or decreasing at the turning instant is not physically precise. \(\textbf{Final answer:}\) The particle is at \(x=+A\) with zero instantaneous speed; it slows down while approaching this instant.

Explanation: The classification must be made using both the direction of the force and its dependence on displacement from equilibrium. For particle P, \(F=-kx\), so the force is directly proportional to displacement and always directed towards \(x=0\). This is the standard linear restoring-force condition for SHM, with angular frequency \(\sqrt{k/m}\). For particle Q, \(F=+kx\). A positive displacement produces positive force, driving the particle farther from \(x=0\), while a negative displacement produces negative force, again driving it away. Therefore, \(x=0\) is an unstable equilibrium, and the motion near it is not SHM. For particle R, \(F=-kx^3/b^2\). The sign of the force is opposite to the sign of displacement, so it is restoring. However, the force magnitude is proportional to \(|x|^3\), not \(|x|\), and the acceleration-displacement ratio is not constant. It can execute bounded oscillations but not exact SHM of finite amplitude. For particle S, \(F=-k(x-b)\). Its equilibrium occurs at \(x=b\), not at the coordinate origin. Measuring displacement from this shifted equilibrium as \(y=x-b\) gives \(F=-ky\), which is the SHM restoring-force law. Thus, a linear restoring force need not vanish at the chosen coordinate origin. The correct classification is P and S as SHM systems, Q as an unstable system, and R as a nonlinear restoring oscillator.

Explanation: \(\textbf{Speed just before collision:}\) The original oscillator has \[ m=1\,\text{kg},\qquad k=100\,\text{N m}^{-1} \] Its angular frequency is \[ \omega=\sqrt{\frac{k}{m}} \] \[ \omega=\sqrt{\frac{100}{1}}=10\,\text{rad s}^{-1} \] The amplitude is \[ A=0.10\,\text{m} \] At equilibrium, the speed is maximum. \[ v_{\max}=A\omega \] \[ v_{\max}=0.10(10)=1\,\text{m s}^{-1} \] The velocity is positive because the block moves right. \(\textbf{Conservation of momentum during collision:}\) The second block is initially at rest. The two blocks stick together, so they acquire a common velocity \(V\). Horizontal momentum is conserved during the short collision. \[ mv_{\max}+m(0)=(2m)V \] Substituting, \[ 1(1)+1(0)=2V \] \[ V=0.50\,\text{m s}^{-1} \] \(\textbf{Changed oscillator after collision:}\) The combined mass is \[ M=2\,\text{kg} \] The spring constant remains \(100\,\text{N m}^{-1}\). Therefore, \[ \omega'=\sqrt{\frac{k}{M}} \] \[ \omega'=\sqrt{\frac{100}{2}} \] \[ \omega'=5\sqrt2\,\text{rad s}^{-1} \] \(\textbf{New amplitude from the collision state:}\) The collision occurs at equilibrium. Immediately after collision, the joined blocks remain at \(x=0\), with speed \(0.50\,\text{m s}^{-1}\). Thus, this speed is the maximum speed of the new oscillation. \[ V=A'\omega' \] Therefore, \[ A'=\frac{V}{\omega'} \] \[ A'=\frac{0.50}{5\sqrt2} \] \[ A'=\frac{0.10}{\sqrt2}\,\text{m} \] \[ A'=5\sqrt2\,\text{cm} \] \(\textbf{New period:}\) \[ T'=\frac{2\pi}{\omega'} \] \[ T'=\frac{2\pi}{5\sqrt2} \] \[ T'=\frac{\pi\sqrt2}{5}\,\text{s} \] \(\textbf{Physical interpretation:}\) Momentum is conserved during the collision, but mechanical energy is not conserved because the collision is perfectly inelastic. The blocks move together after collision, increasing the oscillating mass. The reduced equilibrium-crossing speed and increased mass both affect the new oscillation. The amplitude must therefore be calculated using the post-collision speed and the new angular frequency. \(\textbf{Final answer:}\) \[ \boxed{A'=5\sqrt2\,\text{cm},\qquad T'=\frac{\pi\sqrt2}{5}\,\text{s}} \]

Explanation: \(\textbf{Angular frequency of the unchanged spring system:}\) The mass and spring constant remain unchanged by the impulse. \[ \omega=\sqrt{\frac{k}{m}} \] \[ \omega=\sqrt{\frac{100}{1}} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{State immediately after the impulse:}\) The block was momentarily at rest at its original positive extreme. Its coordinate does not change during the very short impulse. Therefore, \[ x_0=+0.05\,\text{m} \] The imposed velocity is \[ v_0=+0.5\sqrt3\,\text{m s}^{-1} \] \(\textbf{New amplitude from displacement and velocity:}\) For SHM, \[ v^2=\omega^2(A'^2-x^2) \] Rearranging, \[ A'^2=x_0^2+\frac{v_0^2}{\omega^2} \] Substituting, \[ A'^2=(0.05)^2+\frac{(0.5\sqrt3)^2}{10^2} \] \[ A'^2=0.0025+\frac{0.75}{100} \] \[ A'^2=0.0025+0.0075 \] \[ A'^2=0.0100\,\text{m}^2 \] Therefore, \[ A'=0.10\,\text{m}=10\,\text{cm} \] \(\textbf{Selecting the new phase:}\) Let the new time origin be the instant immediately after the impulse. Write \[ x=A'\cos(\omega t+\phi) \] At \(t=0\), \[ \cos\phi=\frac{x_0}{A'} \] \[ \cos\phi=\frac{0.05}{0.10}=\frac12 \] The initial velocity is positive. Since \[ v=-A'\omega\sin(\omega t+\phi) \] we require \(\sin\phi\lt 0\). Hence, the suitable phase constant is \[ \phi=-\frac{\pi}{3} \] \(\textbf{Time to the next positive extreme:}\) The next positive extreme occurs when the phase first reaches zero. \[ \omega t+\phi=0 \] Substituting, \[ 10t-\frac{\pi}{3}=0 \] \[ t=\frac{\pi}{30}\,\text{s} \] \(\textbf{Physical check:}\) The block initially lies at \(+5\,\text{cm}\) and moves farther in the positive direction. The new positive extreme must therefore lie beyond \(+5\,\text{cm}\), which is consistent with an amplitude of \(10\,\text{cm}\). The period does not change because neither the mass nor spring stiffness changes. The impulse changes the amplitude and initial phase by altering the instantaneous velocity. \(\textbf{Final answer:}\) \[ \boxed{A'=10\,\text{cm},\qquad t_{\mathrm{first}}=\frac{\pi}{30}\,\text{s}} \]

Explanation: \(\textbf{Equilibrium of the floating cylinder:}\) At equilibrium, the upward buoyant force balances the cylinder's weight. Let \(V_0\) be the submerged volume at equilibrium. \[ \rho gV_0=mg \] Therefore, \[ \rho V_0=m \] \(\textbf{Change in buoyant force on displacement:}\) Let \(x\) be the downward displacement from equilibrium. A downward movement increases the displaced liquid volume by \[ \Delta V=Sx \] The corresponding increase in upward buoyant force is \[ \Delta F_B=\rho gSx \] Because the equilibrium buoyant force already balances the weight, the additional buoyant force is the net restoring force. Taking the downward direction as positive, \[ F_{\mathrm{net}}=-\rho gSx \] \(\textbf{Effective restoring-force constant:}\) Comparing with \(F=-kx\), \[ k_{\mathrm{eff}}=\rho gS \] The cross-sectional area is \[ S=50\,\text{cm}^2 \] \[ S=50\times10^{-4}=0.005\,\text{m}^2 \] Therefore, \[ k_{\mathrm{eff}}=1000(10)(0.005) \] \[ k_{\mathrm{eff}}=50\,\text{N m}^{-1} \] \(\textbf{Angular frequency:}\) Newton's second law gives \[ ma=-k_{\mathrm{eff}}x \] Hence, \[ \omega=\sqrt{\frac{k_{\mathrm{eff}}}{m}} \] Substituting, \[ \omega=\sqrt{\frac{50}{0.50}} \] \[ \omega=\sqrt{100} \] \[ \omega=10\,\text{rad s}^{-1} \] \(\textbf{Amplitude and maximum speed:}\) The cylinder is released from rest after a downward displacement of \(4\,\text{cm}\). Therefore, \[ A=0.04\,\text{m} \] The maximum speed is \[ v_{\max}=A\omega \] \[ v_{\max}=0.04(10) \] \[ v_{\max}=0.40\,\text{m s}^{-1} \] \(\textbf{Maximum restoring force:}\) \[ F_{\max}=k_{\mathrm{eff}}A \] \[ F_{\max}=50(0.04) \] \[ F_{\max}=2\,\text{N} \] \(\textbf{Physical interpretation:}\) A downward displacement increases the buoyant force, producing an upward restoring force. An upward displacement reduces the buoyant force, producing a downward net restoring force. For a constant cross-section, this force is proportional to displacement, so the float executes SHM under the stated ideal conditions. The restoring force arises from changes in buoyancy rather than from a mechanical spring. \(\textbf{Final answer:}\) \[ \boxed{\omega=10\,\text{rad s}^{-1},\qquad v_{\max}=0.40\,\text{m s}^{-1},\qquad F_{\max}=2\,\text{N}} \]

Explanation: \(\textbf{Relating spring length to applied load:}\) Let the natural length of the spring be \(L_0\). For a suspended load of mass \(m\), static equilibrium gives \[ k(L-L_0)=mg \] Therefore, \[ L=L_0+\frac{mg}{k} \] \(\textbf{Using the first measurement:}\) For \(m_1=0.20\,\text{kg}\), \[ L_1=0.14\,\text{m} \] Thus, \[ 0.14=L_0+\frac{0.20(10)}{k} \] \[ 0.14=L_0+\frac{2}{k} \] \(\textbf{Using the second measurement:}\) For \(m_2=0.50\,\text{kg}\), \[ L_2=0.20\,\text{m} \] Therefore, \[ 0.20=L_0+\frac{0.50(10)}{k} \] \[ 0.20=L_0+\frac{5}{k} \] \(\textbf{Eliminating the unknown natural length:}\) Subtracting the first relation from the second, \[ 0.20-0.14=\frac{5-2}{k} \] \[ 0.06=\frac3k \] Therefore, \[ k=\frac{3}{0.06} \] \[ k=50\,\text{N m}^{-1} \] \(\textbf{Finding the natural length:}\) Substitute the calculated spring constant into the first measurement equation. \[ 0.14=L_0+\frac{2}{50} \] \[ 0.14=L_0+0.04 \] \[ L_0=0.10\,\text{m} \] \[ L_0=10\,\text{cm} \] \(\textbf{Period with a }2.0\,\textbf{kg load:}\) The period of an ideal vertical spring-block oscillator is \[ T=2\pi\sqrt{\frac{m}{k}} \] Substituting, \[ T=2\pi\sqrt{\frac{2.0}{50}} \] \[ T=2\pi\sqrt{0.04} \] \[ T=2\pi(0.20) \] \[ T=\frac{2\pi}{5}\,\text{s} \] \(\textbf{Physical interpretation:}\) The increase in spring length between two different loads determines stiffness without first knowing the natural length. The natural length is then obtained by subtracting the static extension from either measured length. When the load is changed, the equilibrium extension changes, but the natural frequency depends on the load mass and spring stiffness rather than on the absolute spring length. The calculated force constant has unit \(\text{N m}^{-1}\), consistent with force divided by extension. \(\textbf{Final answer:}\) \[ \boxed{L_0=10\,\text{cm},\qquad k=50\,\text{N m}^{-1},\qquad T=\frac{2\pi}{5}\,\text{s}} \]

Explanation: \(\textbf{Equilibrium extension of the spring:}\) The equilibrium condition is \[ kl_0=mg \] Therefore, \[ l_0=\frac{mg}{k} \] Substituting, \[ l_0=\frac{0.50(10)}{100} \] \[ l_0=0.05\,\text{m}=5\,\text{cm} \] \(\textbf{Choosing a displacement coordinate:}\) Let downward displacement from equilibrium be positive. The displacement can be written as \[ x=A\cos(\omega t) \] where \[ A=10\,\text{cm} \] The instantaneous spring extension relative to natural length is \[ l=l_0+x \] \(\textbf{Condition for compression:}\) The spring is compressed when its extension is negative. Therefore, \[ l_0+x\lt 0 \] \[ x\lt -l_0 \] Substituting, \[ x\lt -5\,\text{cm} \] Using \(x=10\cos(\omega t)\) in centimetres, \[ 10\cos(\omega t)\lt -5 \] \[ \cos(\omega t)\lt -\frac12 \] \(\textbf{Determining the compressed-phase interval:}\) Over one complete cycle, the inequality holds for \[ \frac{2\pi}{3}\lt \omega t\lt \frac{4\pi}{3} \] The angular width is \[ \Delta\theta=\frac{4\pi}{3}-\frac{2\pi}{3} \] \[ \Delta\theta=\frac{2\pi}{3} \] Thus, the fraction of the period during which the spring is compressed is \[ \eta=\frac{\Delta\theta}{2\pi} \] \[ \eta=\frac{2\pi/3}{2\pi} \] \[ \eta=\frac13 \] \(\textbf{Spring force at the upper extreme:}\) At the upper extreme, \[ x=-10\,\text{cm} \] The extension relative to natural length is \[ l=5-10 \] \[ l=-5\,\text{cm} \] The spring is compressed by \(5\,\text{cm}\). A compressed vertical spring exerts a downward force on the block in this arrangement. Its magnitude is \[ F_s=k|l| \] \[ F_s=100(0.05) \] \[ F_s=5\,\text{N} \] \(\textbf{Net force at the upper extreme:}\) The block's weight is also downward. \[ mg=0.50(10)=5\,\text{N} \] Therefore, the net force is downward with magnitude \[ F_{\mathrm{net}}=5+5=10\,\text{N} \] \(\textbf{Physical interpretation:}\) A spring attached vertically above the block need not exert an upward force throughout the oscillation. If the block rises beyond the spring's natural length, the compressed spring pushes it downward. The net restoring force remains directed towards equilibrium. At the upper extreme, both gravity and spring force act downward. The spring is compressed for only one-third of the cycle because the equilibrium extension is smaller than the oscillation amplitude. \(\textbf{Final answer:}\) The spring is compressed for one-third of each period. At the upper extreme, both the spring force and the net force are directed downward.