Online Test — Oscillations
18 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 18
Simple harmonic motion is characterised by the relation:
$a=\omega^2 x$
$a=-\omega^2 x$
$a=-\omega x$
$a=\omega x$
Explanation: In SHM acceleration is proportional to displacement and directed towards the mean position: $a=-\omega^2 x$.
Question 2 of 18
The displacement in SHM is $x=A\sin(\omega t+\phi)$. The quantity $\phi$ is called the:
amplitude
angular frequency
initial phase (epoch)
time period
Explanation: $\phi$ is the initial phase or epoch, the phase at $t=0$.
Question 3 of 18
The velocity of a particle in SHM is maximum at:
the extreme positions
the mean position
$x=A/2$
$x=A/\sqrt{2}$
Explanation: $v=\omega\sqrt{A^2-x^2}$ is greatest when $x=0$.
Question 4 of 18
The angular frequency $\omega$ is related to the period $T$ by:
$\omega=2\pi T$
$\omega=\frac{2\pi}{T}$
$\omega=\frac{T}{2\pi}$
$\omega=\pi T$
Explanation: $\omega=\frac{2\pi}{T}=2\pi f$.
Question 5 of 18
In SHM the acceleration is maximum at:
the mean position
the extreme positions
all positions
$x=A/2$
Explanation: $a=-\omega^2 x$ is largest in magnitude when $|x|=A$.
Question 6 of 18
The maximum velocity of a particle of amplitude $A$ and angular frequency $\omega$ in SHM is:
$A\omega^2$
$A\omega$
$\frac{A}{\omega}$
$\omega^2$
Explanation: $v_{max}=A\omega$ occurs at the mean position.
Question 7 of 18
The kinetic energy of an SHM particle at displacement $x$ is:
$\frac{1}{2}m\omega^2 x^2$
$\frac{1}{2}m\omega^2(A^2-x^2)$
$\frac{1}{2}m\omega^2 A^2$
$\frac{1}{2}m\omega^2(A^2+x^2)$
Explanation: $KE=\frac{1}{2}mv^2=\frac{1}{2}m\omega^2(A^2-x^2)$.
Question 8 of 18
The potential energy of a particle in SHM at displacement $x$ is:
$\frac{1}{2}m\omega^2 x^2$
$\frac{1}{2}m\omega^2 A^2$
$\frac{1}{2}m\omega^2(A^2-x^2)$
zero
Explanation: $PE=\frac{1}{2}m\omega^2 x^2=\frac{1}{2}kx^2$.
Question 9 of 18
The total mechanical energy of a particle in SHM is:
$\frac{1}{2}m\omega^2 x^2$
$\frac{1}{2}m\omega^2 A^2$
zero at the extremes
maximum at the mean position
Explanation: $E=KE+PE=\frac{1}{2}m\omega^2 A^2$ and is constant.
Question 10 of 18
KE equals PE in SHM at a displacement of:
$A$
$\frac{A}{2}$
$\frac{A}{\sqrt{2}}$
$0$
Explanation: Setting $KE=PE$ gives $A^2=2x^2$, so $x=\frac{A}{\sqrt{2}}$.
Question 11 of 18
The total energy of an SHM is proportional to:
$A$
$A^2$
$\frac{1}{A}$
$A^3$
Explanation: $E=\frac{1}{2}m\omega^2 A^2\propto A^2$.
Question 12 of 18
The time period of a spring–mass system is:
$2\pi\sqrt{\frac{k}{m}}$
$2\pi\sqrt{\frac{m}{k}}$
$2\pi\sqrt{\frac{L}{g}}$
$2\pi\sqrt{mk}$
Explanation: $T=2\pi\sqrt{\frac{m}{k}}$ from $\omega=\sqrt{k/m}$.
Question 13 of 18
The time period of a simple pendulum of length $L$ is:
$2\pi\sqrt{\frac{g}{L}}$
$2\pi\sqrt{\frac{L}{g}}$
$2\pi\sqrt{Lg}$
$2\pi\frac{L}{g}$
Explanation: $T=2\pi\sqrt{\frac{L}{g}}$ for small oscillations.
Question 14 of 18
A simple pendulum is taken to the Moon ($g$ smaller). Its time period will:
decrease
increase
stay the same
become zero
Explanation: Since $T\propto\frac{1}{\sqrt{g}}$, a smaller $g$ gives a larger period.
Question 15 of 18
Two identical springs of constant $k$ joined in series have an effective constant of:
$2k$
$k$
$\frac{k}{2}$
$4k$
Explanation: $\frac{1}{k_s}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}$, so $k_s=\frac{k}{2}$.
Question 16 of 18
In a damped oscillation, the amplitude:
remains constant
grows linearly
decreases exponentially with time
oscillates between two fixed values
Explanation: The amplitude decays as $A(t)=A_0 e^{-bt/2m}$.
Question 17 of 18
Resonance occurs when the frequency of the driving force is:
twice the natural frequency
half the natural frequency
equal to the natural frequency
zero
Explanation: Amplitude is maximum when the driving frequency equals the natural frequency.
Question 18 of 18
A body in SHM has period $T$. Its kinetic energy varies with a period of:
$T$
$\frac{T}{2}$
$2T$
$4T$
Explanation: KE (and PE) vary at twice the frequency of the motion, i.e. with period $\frac{T}{2}$.