401. A simple pendulum has period \(T\) in a stationary lift. The lift then accelerates downward with acceleration \(\frac{g}{4}\). The new period is
ⓐ. \(\frac{\sqrt{3}}{2}T\)
ⓑ. \(\frac{2}{\sqrt{3}}T\)
ⓒ. \(\frac{T}{2}\)
ⓓ. \(2T\)
Correct Answer: \(\frac{2}{\sqrt{3}}T\)
Explanation: In a stationary lift, the effective acceleration is \(g\).
For downward acceleration \(a=\frac{g}{4}\):
\[
g_{\text{eff}}=g-a
\]
\[
g_{\text{eff}}=g-\frac{g}{4}=\frac{3g}{4}
\]
For the same pendulum length:
\[
T\propto \frac{1}{\sqrt{g_{\text{eff}}}}
\]
Thus,
\[
\frac{T'}{T}=\sqrt{\frac{g}{g_{\text{eff}}}}
\]
\[
\frac{T'}{T}=\sqrt{\frac{g}{3g/4}}
\]
\[
\frac{T'}{T}=\sqrt{\frac{4}{3}}
\]
\[
T'=\frac{2}{\sqrt{3}}T
\]
The downward acceleration reduces effective gravity, so the period increases.
\( \textbf{Final answer:} \) \(T'=\frac{2}{\sqrt{3}}T\).
402. A physical pendulum has the same mass and the same centre-of-mass distance \(d\) from the suspension axis in two cases, but its moment of inertia about the axis is larger in Case 2. Its period in Case 2 is
ⓐ. smaller because the restoring torque is larger
ⓑ. larger because rotational inertia is larger
ⓒ. unchanged because \(I\) is absent from the formula
ⓓ. zero because an extended body cannot oscillate
Correct Answer: larger because rotational inertia is larger
Explanation: For a physical pendulum, the small-oscillation period is \(T=2\pi\sqrt{\frac{I}{mgd}}\). If \(m\), \(g\), and \(d\) remain the same, the restoring torque factor \(mgd\) is unchanged. Increasing \(I\) increases the rotational inertia about the suspension axis. Since \(I\) is in the numerator inside the square root, the period becomes larger. The body oscillates more slowly when it is harder to rotate about the axis.
403. A lightly damped oscillator has an \(x-t\) graph whose successive positive peaks decrease in height, but the time gaps between successive peaks remain nearly equal. This graph shows that
ⓐ. amplitude decreases while period is nearly constant
ⓑ. amplitude remains constant while period decreases
ⓒ. mechanical energy increases after every cycle
ⓓ. damping removes the restoring tendency completely
Correct Answer: amplitude decreases while period is nearly constant
Explanation: In light damping, resistive forces remove mechanical energy slowly. The amplitude therefore decreases from one cycle to the next. However, for weak damping, the time period is often approximately the same as that of the corresponding undamped oscillator. On the graph, decreasing peak heights show amplitude loss, while nearly equal peak spacing shows nearly constant period. The restoring tendency is still present because the system continues to oscillate about the mean position.
404. A damping force is modeled as \(F_d=-bv\), where \(b\) is positive. If the oscillator is moving in the positive direction, the damping force is
ⓐ. in the positive direction
ⓑ. in the negative direction
ⓒ. zero because displacement may be zero
ⓓ. always equal to the restoring force
Correct Answer: in the negative direction
Explanation: In the model \(F_d=-bv\), the damping force is proportional to velocity and opposite in direction. If the oscillator is moving in the positive direction, then \(v\gt0\). Substituting into \(F_d=-bv\) gives a negative damping force. This negative force opposes the motion and removes mechanical energy from the oscillator. The damping force is tied to velocity, so it can be nonzero even when displacement is zero.
405. A forced oscillator is first released and then driven by a periodic force. After a long time, the transient free part becomes negligible. The frequency of the steady motion is then mainly
ⓐ. zero because damping removes all motion instantly
ⓑ. the initial release frequency only
ⓒ. the driving frequency
ⓓ. the amplitude divided by the mass
Correct Answer: the driving frequency
Explanation: Forced oscillation contains a transient part and a steady-state part. The transient part depends on the initial conditions and the natural response of the system. Damping gradually removes this transient contribution. After sufficient time, the sustained motion follows the external periodic driver. The natural frequency still affects the amplitude response, especially near resonance, but the steady oscillation frequency is set by the driving force.
406. Two identical spring oscillators are driven by the same periodic force at their resonance frequency. Oscillator P has light damping, while oscillator Q has heavy damping. The steady amplitude of P is larger mainly because
ⓐ. light damping allows the supplied energy to build a larger response
ⓑ. heavy damping makes the natural frequency exactly zero
ⓒ. light damping removes the restoring force from oscillator P
ⓓ. resonance can occur only when damping is completely absent
Correct Answer: light damping allows the supplied energy to build a larger response
Explanation: At resonance, the driving force transfers energy efficiently to the oscillator. In the steady state, the amplitude is limited by damping because damping dissipates mechanical energy. Light damping dissipates energy less strongly, so the energy supplied by the driver can maintain a larger oscillation amplitude. Heavy damping removes energy more rapidly and therefore keeps the resonance peak lower and broader. The restoring force is still present in both oscillators, and resonance does not require damping to be exactly zero. Therefore, oscillator P has the larger steady amplitude because light damping limits the response less strongly.
407. Two tuning forks of frequencies \(256\,\text{Hz}\) and \(258\,\text{Hz}\) are sounded together. The slow rise and fall of loudness is associated with
ⓐ. critical damping only
ⓑ. static equilibrium
ⓒ. beats
ⓓ. uniform circular motion only
Correct Answer: beats
Explanation: Beats occur when two oscillations of slightly different frequencies are superposed. Their phases gradually shift relative to each other, so the resultant sound alternately becomes louder and softer. The beat frequency is related to the difference between the two frequencies. Here the frequencies are close, so a slow variation in loudness can be heard. This idea belongs near oscillations and sound, but it should not be confused with resonance, which involves driving near a natural frequency.
408. Beats are related to oscillations because they involve
ⓐ. loss of all mechanical energy in one cycle
ⓑ. a single oscillator driven exactly at its natural frequency only
ⓒ. a restoring force proportional to \(x^3\)
ⓓ. superposition of two nearby-frequency oscillations
Correct Answer: superposition of two nearby-frequency oscillations
Explanation: Beats arise when two oscillations with nearly equal frequencies combine. The resultant amplitude changes slowly because the two oscillations alternately reinforce and partially cancel each other. This is different from ordinary free SHM of one oscillator. It is also different from resonance, where an external driver acts near the natural frequency of a system. Beats are therefore a useful nearby concept, but a full treatment belongs more naturally with waves and sound.
409. A mixed oscillator record says: “The amplitude of a spring oscillator is doubled, and its mass is also doubled while the same spring is used.” The correct comparison with the original system is
ⓐ. \(T'=T,\ E'=2E\)
ⓑ. \(T'=2T,\ E'=2E\)
ⓒ. \(T'=\sqrt{2}T,\ E'=4E\)
ⓓ. \(T'=2\sqrt{2}T,\ E'=E\)
Correct Answer: \(T'=\sqrt{2}T,\ E'=4E\)
Explanation: For a spring oscillator:
\[
T=2\pi\sqrt{\frac{m}{k}}
\]
If \(m\) is doubled and \(k\) is unchanged:
\[
T'=2\pi\sqrt{\frac{2m}{k}}
\]
\[
T'=\sqrt{2}T
\]
Total energy is
\[
E=\frac{1}{2}kA^2
\]
If amplitude is doubled:
\[
E'=\frac{1}{2}k(2A)^2
\]
\[
E'=4E
\]
The mass change affects the period, while the amplitude change affects the stored energy for the same spring.
\( \textbf{Final answer:} \) \(T'=\sqrt{2}T\), \(E'=4E\).
410. The table lists four claims about oscillations.
| Claim | Statement |
|---|
| I | Every SHM is oscillatory, but every oscillatory motion need not be SHM. |
| II | In ideal SHM, velocity and acceleration are always in the same direction. |
| III | The small-angle pendulum formula requires \(\theta\) to be small in radians. |
| IV | A real resonance peak is limited by damping. |
The valid claims are
ⓐ. II, III, and IV only
ⓑ. I and II only
ⓒ. I, III, and IV only
ⓓ. I, II, III, and IV
Correct Answer: I, III, and IV only
Explanation: Claim I is valid because SHM is a special kind of oscillatory motion with a linear restoring condition. Claim II is not valid because velocity and acceleration can have different directions; for example, while moving away from the mean position, acceleration is toward the mean position. Claim III is valid because \(\sin\theta\approx\theta\) works only for small angles measured in radians. Claim IV is valid because damping dissipates energy and prevents real resonance amplitudes from growing without limit. The set combines classification, phase-direction, approximation, and resonance conditions.
411. In an ideal spring oscillator, \(x=A\cos\omega t\). The potential energy varies with time as
ⓐ. \(U=\frac{1}{2}kA\cos^2\omega t\)
ⓑ. \(U=\frac{1}{2}kA^2\cos\omega t\)
ⓒ. \(U=\frac{1}{2}kA^2\cos^2\omega t\)
ⓓ. \(U=\frac{1}{2}mA^2\sin^2\omega t\)
Correct Answer: \(U=\frac{1}{2}kA^2\cos^2\omega t\)
Explanation: For a spring oscillator, the potential energy at displacement \(x\) is \(U=\frac{1}{2}kx^2\). Given \(x=A\cos\omega t\), substitute this displacement into the energy expression. This gives \(U=\frac{1}{2}k(A\cos\omega t)^2\). Squaring the displacement gives \(U=\frac{1}{2}kA^2\cos^2\omega t\). The square is essential because spring potential energy is non-negative for both positive and negative displacements. A direct \(\cos\omega t\) dependence would incorrectly make potential energy negative during part of the cycle.
412. For an ideal spring oscillator with \(x=A\cos\omega t\), the kinetic energy varies with time as
ⓐ. \(K=\frac{1}{2}kA^2\sin\omega t\)
ⓑ. \(K=\frac{1}{2}kA^2\cos^2\omega t\)
ⓒ. \(K=\frac{1}{2}kA^2\sin^2\omega t\)
ⓓ. \(K=\frac{1}{2}mA^2\cos^2\omega t\)
Correct Answer: \(K=\frac{1}{2}kA^2\sin^2\omega t\)
Explanation: \( \textbf{Given displacement:} \)
\[
x=A\cos\omega t
\]
Velocity is
\[
v=\frac{dx}{dt}=-A\omega\sin\omega t
\]
Kinetic energy is
\[
K=\frac{1}{2}mv^2
\]
Substitute \(v=-A\omega\sin\omega t\):
\[
K=\frac{1}{2}mA^2\omega^2\sin^2\omega t
\]
For a spring oscillator,
\[
\omega^2=\frac{k}{m}
\]
So,
\[
m\omega^2=k
\]
Therefore,
\[
K=\frac{1}{2}kA^2\sin^2\omega t
\]
This expression is maximum at the mean position and zero at the extreme positions.
\( \textbf{Final answer:} \) \(K=\frac{1}{2}kA^2\sin^2\omega t\).
413. The kinetic and potential energies of an ideal spring oscillator repeat their values after every
ⓐ. \(T\)
ⓑ. \(\frac{T}{2}\)
ⓒ. \(\frac{T}{4}\)
ⓓ. \(2T\)
Correct Answer: \(\frac{T}{2}\)
Explanation: In spring SHM, potential energy depends on \(x^2\), and kinetic energy depends on \(v^2\). Because of the square, the energy values are the same at opposite displacements of equal magnitude. For example, \(x=+A\) and \(x=-A\) both give maximum potential energy and zero kinetic energy. These opposite extreme positions occur half a period apart. Therefore, each separate energy form repeats after \(\frac{T}{2}\), even though the displacement itself repeats after \(T\).
414. A spring oscillator has total energy \(E\). At a certain instant, its potential energy is equal to its kinetic energy. The speed at that instant is
ⓐ. \(\frac{v_{\max}}{2}\)
ⓑ. \(\frac{v_{\max}}{\sqrt{2}}\)
ⓒ. \(\frac{\sqrt{3}}{2}v_{\max}\)
ⓓ. \(v_{\max}\)
Correct Answer: \(\frac{v_{\max}}{\sqrt{2}}\)
Explanation: \( \textbf{Energy condition:} \) \(K=U\).
Since
\[
E=K+U
\]
equal kinetic and potential energies mean
\[
K=\frac{E}{2}
\]
At the mean position, all energy is kinetic:
\[
E=\frac{1}{2}mv_{\max}^2
\]
At the required instant:
\[
K=\frac{1}{2}mv^2
\]
Use \(K=\frac{E}{2}\):
\[
\frac{1}{2}mv^2=\frac{1}{2}\left(\frac{1}{2}mv_{\max}^2\right)
\]
Cancel \(\frac{1}{2}m\):
\[
v^2=\frac{v_{\max}^2}{2}
\]
Taking the positive value for speed:
\[
v=\frac{v_{\max}}{\sqrt{2}}
\]
The speed is not half the maximum speed because kinetic energy depends on \(v^2\).
\( \textbf{Final answer:} \) \(v=\frac{v_{\max}}{\sqrt{2}}\).
415. Use the graph description below.
For an ideal spring oscillator, \(U(t)\) and \(K(t)\) are plotted against time. Both curves are always non-negative, and whenever one curve is maximum, the other is zero. Their sum is a horizontal line.
The correct interpretation of the graph is
ⓐ. total energy stays constant as \(K\) and \(U\) exchange
ⓑ. total energy drops to zero when either curve is maximum
ⓒ. \(K\) and \(U\) remain equal at all instants in the cycle
ⓓ. displacement stays constant while energies exchange
Correct Answer: total energy stays constant as \(K\) and \(U\) exchange
Explanation: In ideal spring SHM, kinetic energy and potential energy vary with time. When the oscillator is at an extreme, potential energy is maximum and kinetic energy is zero. When it passes through the mean position, kinetic energy is maximum and spring potential energy is minimum. The horizontal sum shows that total mechanical energy remains constant. The graph therefore represents energy exchange without loss, not disappearance of energy.
416. A spring oscillator has \(k=80\,\text{N m}^{-1}\) and amplitude \(0.20\,\text{m}\). At the instant when kinetic energy equals potential energy, the magnitude of displacement is
ⓐ. \(0.050\,\text{m}\)
ⓑ. \(0.10\,\text{m}\)
ⓒ. \(0.141\,\text{m}\)
ⓓ. \(0.20\,\text{m}\)
Correct Answer: \(0.141\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(A=0.20\,\text{m}\).
When kinetic energy equals potential energy:
\[
U=\frac{E}{2}
\]
For a spring oscillator:
\[
U=\frac{1}{2}kx^2
\]
and
\[
E=\frac{1}{2}kA^2
\]
So,
\[
\frac{U}{E}=\frac{x^2}{A^2}
\]
Set \(\frac{U}{E}=\frac{1}{2}\):
\[
\frac{x^2}{A^2}=\frac{1}{2}
\]
\[
|x|=\frac{A}{\sqrt{2}}
\]
Substitute \(A=0.20\,\text{m}\):
\[
|x|=\frac{0.20}{\sqrt{2}}
\]
\[
|x|\approx0.141\,\text{m}
\]
The value of \(k\) cancels because the question asks for a position as a fraction of amplitude.
\( \textbf{Final answer:} \) \(|x|\approx0.141\,\text{m}\).
417. A SHM particle has \(x=A\sin\omega t\). At \(t=\frac{T}{8}\), the fraction of total spring energy stored as potential energy is
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(\frac{3}{4}\)
ⓓ. \(1\)
Correct Answer: \(\frac{1}{2}\)
Explanation: \( \textbf{Given displacement:} \)
\[
x=A\sin\omega t
\]
At \(t=\frac{T}{8}\), use \(\omega T=2\pi\):
\[
\omega t=\omega\left(\frac{T}{8}\right)=\frac{2\pi}{8}
\]
\[
\omega t=\frac{\pi}{4}
\]
So,
\[
x=A\sin\frac{\pi}{4}
\]
\[
x=\frac{A}{\sqrt{2}}
\]
For spring SHM:
\[
\frac{U}{E}=\frac{x^2}{A^2}
\]
Substitute:
\[
\frac{U}{E}=\frac{\left(\frac{A}{\sqrt{2}}\right)^2}{A^2}
\]
\[
\frac{U}{E}=\frac{1}{2}
\]
At one-eighth of a period after starting from the mean position, energy is equally divided between kinetic and potential forms.
\( \textbf{Final answer:} \) \(\frac{U}{E}=\frac{1}{2}\).
418. In the displacement equation \(x=A\sin(\omega t+\phi)\), changing only \(\phi\) while keeping \(A\) and \(\omega\) fixed changes
ⓐ. the starting state of the oscillator
ⓑ. the amplitude of the oscillator
ⓒ. the angular frequency of the oscillator
ⓓ. the total energy of a spring oscillator
Correct Answer: the starting state of the oscillator
Explanation: The constant \(\phi\) is the initial phase. It decides the phase of the oscillator at \(t=0\). Changing \(\phi\) can change the initial displacement and the initial direction of motion. It does not change the amplitude \(A\), which fixes the maximum displacement. It also does not change \(\omega\), which fixes the time period. For the same spring oscillator and same amplitude, changing only phase does not change the total energy.
419. Two SHM equations are \(x_1=A\sin\omega t\) and \(x_2=A\cos\omega t\). The phase difference between them is
ⓐ. \(\frac{0\pi}{2}\,\text{rad}\)
ⓑ. \(\frac{\pi}{2}\,\text{rad}\)
ⓒ. \(\frac{2\pi}{2}\,\text{rad}\)
ⓓ. \(\frac{4\pi}{2}\,\text{rad}\)
Correct Answer: \(\frac{\pi}{2}\,\text{rad}\)
Explanation: The cosine function can be written as a sine function with a phase shift:
\[
\cos\omega t=\sin\left(\omega t+\frac{\pi}{2}\right)
\]
Therefore,
\[
x_2=A\sin\left(\omega t+\frac{\pi}{2}\right)
\]
while
\[
x_1=A\sin\omega t
\]
The phase difference is \(\frac{\pi}{2}\). This means one motion is a quarter cycle ahead of the other if they have the same amplitude and angular frequency. The two equations do not describe different kinds of motion; they describe the same SHM form with different starting phases.
\( \textbf{Final answer:} \) Phase difference \(=\frac{\pi}{2}\).
420. A laboratory graph of \(a\) versus \(x\) for a supposed SHM has slope \(-49\,\text{s}^{-2}\). The time period of the motion is
ⓐ. \(\frac{2\pi}{7}\,\text{s}\)
ⓑ. \(7\,\text{s}\)
ⓒ. \(14\pi\,\text{s}\)
ⓓ. \(\frac{7}{2\pi}\,\text{s}\)
Correct Answer: \(\frac{2\pi}{7}\,\text{s}\)
Explanation: \( \textbf{Given graph information:} \) The slope of the \(a-x\) graph is \(-49\,\text{s}^{-2}\).
For SHM:
\[
a=-\omega^2x
\]
So the slope of the \(a-x\) graph is
\[
-\omega^2
\]
Therefore,
\[
-\omega^2=-49\,\text{s}^{-2}
\]
\[
\omega^2=49\,\text{s}^{-2}
\]
\[
\omega=7\,\text{rad s}^{-1}
\]
Time period is
\[
T=\frac{2\pi}{\omega}
\]
\[
T=\frac{2\pi}{7}\,\text{s}
\]
The slope gives \(\omega^2\), not \(T\) directly.
\( \textbf{Final answer:} \) \(T=\frac{2\pi}{7}\,\text{s}\).
