Pravegaa PYQ Series · Volume 2: Classical Mechanics & Special Relativity

CM10: Oscillations and Normal Modes

67 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

IIT JAM: 18
GATE: 5
CSIR NET: 17
JEST: 16
TIFR GS: 11

CM10.1IIT JAM 2026, Q21MCQ | 2 marks

A mass attached to the bottom end of a vertical massless spring stretches the spring by \(\Delta x\). The system executes oscillation with a time period \(T=0.2\,\text{s}\). The value of \(\Delta x\), in cm, rounded off to the nearest integer is:

[Assume the acceleration due to gravity \(g=9.8\,\text{m/s}^2\)]

Show solution

At equilibrium \(k\Delta x=mg\), so \(k/m=g/\Delta x\) and \[ T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{\Delta x}{g}}\quad\Longrightarrow\quad \Delta x=g\left(\frac{T}{2\pi}\right)^2 . \] \[ \Delta x=9.8\times\left(\frac{0.2}{2\pi}\right)^2=9.8\times1.013\times10^{-3}=9.93\times10^{-3}\ \text{m}\approx1\ \text{cm} \] \[ \boxed{\Delta x\approx1\ \text{cm}} \]

Answer: (D) \(1\)

CM10.2IIT JAM 2026, Q51NAT | 2 marks

Similar: CM10.21 (GATE 2024)

A particle of mass \(m\) undergoes periodic motion in one-dimension with its total energy given as \(E=\frac12m\dot{x}^2+\frac14kx^4\), where \(k\) is a positive constant and \(\dot{x}=\dfrac{dx}{dt}\). Assuming that \(E\) is conserved, the time period \(T\) has the relation \(T\propto E^{-1/n}\). The value of \(n\) is . (Answer in integer)

Show solution

The amplitude \(A\) is fixed by \(E=\tfrac14kA^4\), so \(A=(4E/k)^{1/4}\). The period is \[ T=4\int_0^A\frac{dx}{\dot x}=4\int_0^A\frac{dx}{\sqrt{\dfrac{2}{m}\left(E-\dfrac14kx^4\right)}} . \] Substituting \(x=Au\) and \(E=\tfrac14kA^4\): \[ T=4\int_0^1\frac{A\,du}{\sqrt{\dfrac{k A^4}{2m}(1-u^4)}}=\frac{4}{A}\sqrt{\frac{2m}{k}}\int_0^1\frac{du}{\sqrt{1-u^4}}\ \propto\ \frac1A\ \propto\ E^{-1/4}. \] Hence \(T\propto E^{-1/4}\) and \(n=4\). \[ \boxed{n=4} \]

Answer: \(4\)

CM10.3IIT JAM 2025, Q20MCQ | 2 marks

Consider the superposition of two orthogonal simple harmonic motions \(y_1=a\cos2\omega t\) and \(y_2=b\cos(\omega t+\phi)\). If \(\phi=\pi\), the resultant motion will represent (\(a\), \(b\) and \(\omega\) are constants with appropriate dimensions)

a parabola
a hyperbola
an ellipse
a circle

Show solution

With \(\phi=\pi\), \(y_2=-b\cos\omega t\), so \(\cos\omega t=-y_2/b\). Using \(\cos2\omega t=2\cos^2\omega t-1\), \[y_1=a\left(2\cos^2\omega t-1\right)=a\left(\frac{2y_2^2}{b^2}-1\right)=\frac{2a}{b^2}\,y_2^2-a .\] This is linear in \(y_1\) and quadratic in \(y_2\): the path is a \(\boxed{\text{parabola}}\) (the arc with \(|y_2|\le b\)).

Answer: (A) a parabola

CM10.4IIT JAM 2024, Q33MSQ | 2 marks

A spring-mass system (spring constant \(80\ \mathrm{N/m}\) and damping coefficient \(40\ \mathrm{N\text{-}s/m}\)), initially at rest, is lying along the y-axis in the horizontal plane. One end of the spring is fixed and the mass (\(5\ \mathrm{kg}\)) is attached at its other end. The mass is pulled along the y-axis by \(0.5\ \mathrm{m}\) from its equilibrium position and then released. Choose the correct statement(s).

(Ignore mass of the spring)

Motion will be under damped
Trajectory of the mass will be \(y(t)=\frac12(1+t)e^{-4t}\)
Motion will be critically damped
Trajectory of the mass will be \(y(t)=\frac12(1+4t)e^{-4t}\)

Show solution

The equation of motion is \(m\ddot y+b\dot y+ky=0\), i.e. \(\ddot y+2\gamma\dot y+\omega_0^2y=0\) with \[ \gamma=\frac{b}{2m}=\frac{40}{2\times5}=4\ \mathrm{s^{-1}},\qquad \omega_0=\sqrt{\frac km}=\sqrt{\frac{80}{5}}=4\ \mathrm{s^{-1}} . \] Since \(\gamma=\omega_0\), the auxiliary equation \(D^2+8D+16=(D+4)^2=0\) has the repeated root \(D=-4\). The motion is critically damped: (C) is true and (A) is false. The general solution is \[ y(t)=(c_1+c_2t)e^{-4t},\qquad \dot y(t)=\left[c_2-4(c_1+c_2t)\right]e^{-4t} . \] Initial conditions: \(y(0)=0.5\Rightarrow c_1=\tfrac12\), and \(\dot y(0)=0\Rightarrow c_2=4c_1=2\). Therefore \[ y(t)=\left(\tfrac12+2t\right)e^{-4t}=\boxed{\tfrac12(1+4t)e^{-4t}} \] so (D) is true and (B) is false.

Answer: (C), (D)

CM10.5IIT JAM 2023, Q23MCQ | 2 marks

A uniform stick of length \(l\) and mass \(m\) pivoted at its top end is oscillating with an angular frequency \(\omega_{\rm r}\). Assuming small oscillations, the ratio \(\omega_{\rm r}/\omega_{\rm s}\), where \(\omega_{\rm s}\) is the angular frequency of a simple pendulum of the same length, will be

\(\sqrt3\)
\(\sqrt{\dfrac32}\)
\(\sqrt2\)
\(\dfrac1{\sqrt3}\)

Show solution

For a physical pendulum, \(\omega_{\rm r}=\sqrt{\dfrac{mgl'}{I}}\), where \(l'\) is the distance of the centre of mass from the pivot and \(I\) is the moment of inertia about the pivot. For a uniform stick pivoted at one end, \(l'=l/2\) and \[ I=\frac{ml^2}{12}+m\left(\frac l2\right)^2=\frac{ml^2}{3}. \] Hence \[ \omega_{\rm r}=\sqrt{\frac{mg\,l/2}{ml^2/3}}=\sqrt{\frac{3g}{2l}},\qquad \omega_{\rm s}=\sqrt{\frac gl} \quad\Rightarrow\quad\frac{\omega_{\rm r}}{\omega_{\rm s}}=\sqrt{\frac32}. \] \[\boxed{\sqrt{3/2}}\]

Answer: (B) \(\sqrt{3/2}\)

CM10.6IIT JAM 2023, Q44NAT | 1 mark

A single pendulum hanging vertically in an elevator has a time period \(T_0\) when the elevator is stationary. If the elevator moves upward with an acceleration of \(a=0.2g\), the time period of oscillations is \(T_1\). Here \(g\) is the acceleration due to gravity. The ratio \(\dfrac{T_0}{T_1}\) is (rounded off to two decimal places).

Show solution

In the upward-accelerating elevator the effective gravity is \(g_{\rm eff}=g+a=1.2g\). With \(T=2\pi\sqrt{l/g_{\rm eff}}\): \[ T_0=2\pi\sqrt{\frac lg},\qquad T_1=2\pi\sqrt{\frac{l}{1.2g}}\quad\Rightarrow\quad\frac{T_0}{T_1}=\sqrt{1.2}=1.095\approx1.10 . \] \[\boxed{T_0/T_1\approx1.10}\]

Answer: 1.10 (range 1.09 to 1.11)

CM10.7IIT JAM 2023, Q47NAT | 1 mark

Consider a spring mass system with mass 0.5 kg and spring constant \(k=2\) N m\(^{-1}\) in a viscous medium with drag coefficient \(b=3\) kg s\(^{-1}\). The additional mass required so that the motion becomes critically damped is kg (rounded off to three decimal places).

Show solution

For \(m\ddot x+b\dot x+kx=0\), critical damping requires \(b^2=4mk\). Hence \[ m_c=\frac{b^2}{4k}=\frac{3^2}{4\times2}=\frac98=1.125\ \text{kg}, \] and the additional mass needed is \[ \Delta m=1.125-0.5=0.625\ \text{kg}. \] \[\boxed{\Delta m=0.625\ \text{kg}}\]

Answer: 0.625 (range 0.620 to 0.630)

CM10.8IIT JAM 2022, Q27MCQ | 2 marks

A particle is subjected to two simple harmonic motions along the \(x\) and \(y\) axes, described by \(x(t)=a\sin(2\omega t+\pi)\) and \(y(t)=2a\sin(\omega t)\). The resultant motion is given by

\(\dfrac{x^2}{a^2}+\dfrac{y^2}{4a^2}=1\)
\(x^2+y^2=1\)
\(y^2=x^2\left(1-\dfrac{x^2}{4a^2}\right)\)
\(x^2=y^2\left(1-\dfrac{y^2}{4a^2}\right)\)

Show solution

From \(y=2a\sin\omega t\), \(\sin\omega t=\dfrac{y}{2a}\). Then \[ x=a\sin(2\omega t+\pi)=-a\sin2\omega t=-2a\sin\omega t\cos\omega t=-y\cos\omega t=-y\sqrt{1-\frac{y^2}{4a^2}} . \] Squaring, \[\boxed{x^2=y^2\left(1-\frac{y^2}{4a^2}\right)}\]

Answer: (D)

CM10.9IIT JAM 2022, Q34MSQ | 2 marks

A particle is executing simple harmonic motion with time period \(T\). Let \(x\), \(v\) and \(a\) denote the displacement, velocity and acceleration of the particle, respectively, at time \(t\). Then,

\(\dfrac{aT}{x}\) does not change with time
\((aT+2\pi v)\) does not change with time
\(x\) and \(v\) are related by an equation of a straight line
\(v\) and \(a\) are related by an equation of an ellipse

Show solution

Let \(x=A\sin(\omega t+\delta)\) with \(\omega=2\pi/T\). Then \[ v=A\omega\cos(\omega t+\delta),\qquad a=-\omega^2A\sin(\omega t+\delta)=-\omega^2x . \]

\(\dfrac{aT}{x}=-\omega^2T=-\dfrac{4\pi^2}{T}\), a constant: (A) true.
\(aT+2\pi v=2\pi A\omega\left[\cos(\omega t+\delta)-\sin(\omega t+\delta)\right]\), which varies with \(t\): (B) false.
\(\dfrac{x^2}{A^2}+\dfrac{v^2}{A^2\omega^2}=1\), an ellipse, not a straight line: (C) false.
\(\dfrac{v^2}{A^2\omega^2}+\dfrac{a^2}{A^2\omega^4}=1\), an ellipse: (D) true.

\[\boxed{\text{(A), (D)}}\]

Answer: (A), (D)

CM10.10IIT JAM 2021, Q34MSQ | 2 marks

Consider the following differential equation that describes the oscillations of a physical system: \[ \alpha\frac{d^2y}{dt^2}+\beta\frac{dy}{dt}+\gamma y=0 \] If \(\alpha\) and \(\beta\) are held fixed, and \(\gamma\) is increased, then,

the frequency of oscillations increases
the oscillations decay faster
the frequency of oscillations decreases
the oscillations decay slower

Show solution

Trying \(y=e^{st}\) gives \(\alpha s^2+\beta s+\gamma=0\), so in the underdamped case \[ s=-\frac{\beta}{2\alpha}\pm i\sqrt{\frac{\gamma}{\alpha}-\frac{\beta^2}{4\alpha^2}},\qquad y=Ae^{-\beta t/2\alpha}\cos\left(\omega' t+\phi\right),\quad \omega'=\sqrt{\frac{\gamma}{\alpha}-\frac{\beta^2}{4\alpha^2}} . \] The decay rate \(\beta/2\alpha\) does not depend on \(\gamma\), so (B) and (D) are false. Increasing \(\gamma\) increases \(\omega'\). \[ \boxed{\text{The frequency increases; the decay rate is unchanged}} \]

Answer: (A)

CM10.11IIT JAM 2020, Q33MSQ | 2 marks

A thin rod of uniform density and length \(2\sqrt3\,m\) is undergoing small oscillations about a pivot point. The time period of oscillation \((T_m)\) is minimum when the distance of the pivot point from the center-of-mass of the rod is \(x_m\). Which of the following is/are correct?

\(x_m=1\,m\)
\(x_m=\dfrac{\sqrt3}{2}\,m\)
\(T_m=\dfrac{2\pi}{\sqrt3}\,s\)
\(T_m=\dfrac{2\pi}{\sqrt5}\,s\)

Show solution

For a rod of length \(l=2\sqrt3\,\text{m}\) pivoted at distance \(x\) from its centre of mass, \(I=\dfrac{Ml^2}{12}+Mx^2\), and \[ T=2\pi\sqrt{\frac{I}{Mgx}}=2\pi\sqrt{\frac{1}{g}\left(\frac{l^2}{12x}+x\right)} \] \(T\) is minimum when \[ \frac{d}{dx}\left(\frac{l^2}{12x}+x\right)=-\frac{l^2}{12x^2}+1=0\ \Rightarrow\ x_m^2=\frac{l^2}{12}=\frac{12}{12}=1\ \Rightarrow\ x_m=1\,\text{m} \] Then, with \(g=10\,\text{m/s}^2\), \[ T_m=2\pi\sqrt{\frac{1}{10}\left(1+1\right)}=2\pi\sqrt{\frac{2}{10}}=\frac{2\pi}{\sqrt5}\,\text{s} \] \[ \boxed{x_m=1\,\text{m},\quad T_m=\frac{2\pi}{\sqrt5}\,\text{s}} \]

Answer: (A), (D)

CM10.12IIT JAM 2020, Q35MSQ | 2 marks

An object executes simple harmonic motion along the \(x\)-direction with angular frequency \(\omega\) and amplitude \(a\). The speed of the object is \(4\,cm/s\) and \(2\,cm/s\) when it is at distances \(2\,cm\) and \(6\,cm\) respectively from the equilibrium position. Which of the following is/are correct?

\(\omega=\sqrt{\dfrac38}\,\text{rad/s}\)
\(\omega=\sqrt{\dfrac56}\,\text{rad/s}\)
\(a=\sqrt{\dfrac{140}{3}}\,\text{cm}\)
\(a=\sqrt{\dfrac{175}{6}}\,\text{cm}\)

Show solution

For \(x=a\sin\omega t\), the speed is \(u=\omega\sqrt{a^2-x^2}\), so \(u^2=\omega^2(a^2-x^2)\): \[ 16=\omega^2(a^2-4)\quad\text{(i)},\qquad 4=\omega^2(a^2-36)\quad\text{(ii)} \] Subtracting, \(12=32\,\omega^2\Rightarrow\omega^2=\dfrac38\). Dividing (i) by (ii), \[ 4=\frac{a^2-4}{a^2-36}\ \Rightarrow\ 3a^2=140\ \Rightarrow\ a=\sqrt{\frac{140}{3}}\,\text{cm} \] \[ \boxed{\omega=\sqrt{\tfrac38}\ \text{rad/s},\quad a=\sqrt{\tfrac{140}{3}}\ \text{cm}} \]

Answer: (A), (C)

CM10.13IIT JAM 2019, Q36MSQ | 2 marks

For an under damped harmonic oscillator with velocity \(v(t)\)

(A)	Rate of energy dissipation varies linearly with \(v(t)\)
(B)	Rate of energy dissipation varies as square of \(v(t)\)
(C)	The reduction in the oscillator frequency, compared to the undamped case, is independent of \(v(t)\)
(D)	For weak damping, the amplitude decays exponentially to zero

Show solution

The equation of motion with a damping force \(-bv\) is \(m\ddot{x}+b\dot{x}+kx=0\). Multiplying by \(\dot{x}=v\), \[ \frac{d}{dt}\left(\frac{1}{2}mv^2+\frac{1}{2}kx^2\right)=-bv^2 \;\Rightarrow\;\frac{dE}{dt}=-bv^2 . \] So the rate of energy dissipation varies as \(v^2\), not linearly: (A) is false and (B) is true.

With \(r=b/2m\) and \(\omega_0^2=k/m\), the under-damped solution is \[ x(t)=Ae^{-rt}\sin(\omega t+\phi),\qquad \omega=\sqrt{\omega_0^2-r^2}. \] The frequency depends only on \(\omega_0\) and \(r\) (i.e. on \(k\), \(m\), \(b\)) and not on \(v(t)\), so (C) is true. The amplitude \(Ae^{-rt}\) decays exponentially to zero, so (D) is true. \[ \boxed{(B),\ (C),\ (D)} \]

Answer: (B), (C), (D)

CM10.14IIT JAM 2018, Q42NAT | 1 mark

A particle of mass \(m\) is moving along the positive \(x\) direction under a potential \(V(x)=\dfrac12kx^2+\dfrac{\lambda}{2x^2}\) (\(k\) and \(\lambda\) are positive constants). If the particle is slightly displaced from its equilibrium position, it oscillates with an angular frequency \((\omega)\) .

(Specify your answer in units of \(\sqrt{\dfrac km}\) as an integer.)

Show solution

\[ \frac{dV}{dx}=kx-\frac{\lambda}{x^3}=0\;\Rightarrow\;x_0^4=\frac{\lambda}{k}\;\Rightarrow\;x_0=\left(\frac{\lambda}{k}\right)^{1/4}\quad(x>0). \] \[ \left.\frac{d^2V}{dx^2}\right|_{x_0}=k+\frac{3\lambda}{x_0^4}=k+3k=4k . \] \[ \omega=\sqrt{\frac{V''(x_0)}{m}}=\sqrt{\frac{4k}{m}}=\boxed{2\sqrt{\frac km}} \]

Answer: 2 (range 2 to 2)

CM10.15IIT JAM 2017, Q19MCQ | 2 marks

A pendulum is made of a massless string of length \(L\) and a small bob of negligible size and mass \(m\). It is released making an angle \(\theta_0\ (\ll1\,\text{rad})\) from the vertical. When passing through the vertical, the string slips a bit from the pivot so that its length increases by a small amount \(\delta\ (\delta\ll L)\) in negligible time. If it swings up to angle \(\theta_1\) on the other side before starting to swing back, then to a good approximation which of the following expressions is correct?

\(\theta_1=\theta_0\)
\(\theta_1=\theta_0\left(1-\dfrac{\delta}{2L}\right)\)
\(\theta_1=\theta_0\left(1-\dfrac{\delta}{L}\right)\)
\(\theta_1=\theta_0\left(1-\dfrac{3\delta}{2L}\right)\)

Show solution

Speed at the lowest point before slipping (small angles): \[ v_0^2=2gL\left(1-\cos\theta_0\right)\approx gL\theta_0^2 . \] During the slip the only force on the bob in the horizontal plane of motion is the string tension, which points towards the pivot; gravity's impulse is negligible in the negligible time. So the angular momentum about the pivot is conserved: \[ mv_0L=mv_1(L+\delta)\;\Rightarrow\;v_1=\frac{v_0L}{L+\delta}. \] After the slip, energy conservation with the new length gives \[ v_1^2=g(L+\delta)\theta_1^2\;\Rightarrow\;\theta_1^2=\frac{v_0^2L^2}{g(L+\delta)^3}=\theta_0^2\left(\frac{L}{L+\delta}\right)^3 . \] \[ \theta_1=\theta_0\left(1+\frac{\delta}{L}\right)^{-3/2}\approx\boxed{\theta_0\left(1-\frac{3\delta}{2L}\right)} \]

Answer: (D) \(\theta_1=\theta_0\left(1-\dfrac{3\delta}{2L}\right)\)

CM10.16IIT JAM 2016, Q9MCQ | 1 mark

Two sinusoidal signals of frequencies \(\omega_x\) and \(\omega_y\) having same amplitude are applied to \(x\)- and \(y\)-channels of a cathode ray oscilloscope (CRO), respectively. The following stationary figure will be observed when

\(\omega_y=\omega_x\)
phase difference is \(0\)
\(\omega_y=2\omega_x\)
phase difference is \(\pi/2\)

Show solution

The figure touches the vertical sides \(x=\pm1\) twice each but the horizontal sides \(y=\pm1\) only once each, so \[ \frac{\omega_x}{\omega_y}=\frac{\text{tangencies with a horizontal side}}{\text{tangencies with a vertical side}}\;\Rightarrow\;\omega_x=2\omega_y . \] This rules out (A) and (C). Take \(y=\sin\omega t\) and \(x=\sin(2\omega t+\phi)\).

For \(\phi=0\): \(x=2\sin\omega t\cos\omega t=2y\sqrt{1-y^2}\), a figure of eight with two lobes, one above and one below the \(x\)-axis, crossing at the origin and reaching \(y=\pm1\) at \(x=0\). This is exactly the figure shown.
For \(\phi=\pi/2\): \(x=\cos2\omega t=1-2y^2\), a parabolic arc, not a closed figure of eight.

\[ \boxed{\text{phase difference }=0} \]

Answer: (B) phase difference is \(0\)

CM10.17IIT JAM 2016, Q25MCQ | 2 marks

A lightly damped harmonic oscillator with natural frequency \(\omega_0\) is driven by a periodic force of frequency \(\omega\). The amplitude of oscillation is maximum when

\(\omega\) is slightly lower than \(\omega_0\)
\(\omega=\omega_0\)
\(\omega\) is slightly higher than \(\omega_0\)
The force is in phase with the displacement

Show solution

For \(\ddot x+2\gamma\dot x+\omega_0^2x=(F_0/m)\cos\omega t\), the steady-state amplitude is \[ A(\omega)=\frac{F_0/m}{\sqrt{\left(\omega_0^2-\omega^2\right)^2+4\gamma^2\omega^2}} . \] Minimising the denominator, \(\dfrac{d}{d\omega^2}\left[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2\right]=0\) gives \[ \omega_r=\sqrt{\omega_0^2-2\gamma^2}<\omega_0 . \] For light damping this is only slightly below \(\omega_0\). (At \(\omega=\omega_0\) it is the velocity, not the displacement, that peaks, and the force there leads the displacement by \(\pi/2\).) \[ \boxed{\omega\ \text{slightly lower than}\ \omega_0} \]

Answer: (A) \(\omega\) is slightly lower than \(\omega_0\)

CM10.18IIT JAM 2016, Q26MCQ | 2 marks

A block of mass \(0.38\,kg\) is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing \(0.02\,kg\) moving with a speed of \(200\,m/s\) hits the block at time \(t=0\) and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by (Spring constant \(=640\,N/m\))

\(2\sin5t\)

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Chapter 10 Oscillations and Normal Modes Previous Year Questions with Solutions

CM10: Oscillations and Normal Modes

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