IMO Practice Test — Oscillations
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
A particle in SHM has displacement $x=A\sin\omega t$. The ratio of its average kinetic energy to average potential energy over one full cycle is:
$1:1$
$2:1$
$1:2$
$3:1$
Explanation: Over a complete cycle the time-averages of KE and PE are each half the total energy, so the ratio is $1:1$.
Question 2 of 12
The displacement of a particle is $x=3\sin\omega t+4\cos\omega t$. The amplitude of the resulting SHM is:
$3$
$4$
$5$
$7$
Explanation: A sum of sine and cosine of the same frequency is SHM of amplitude $\sqrt{3^2+4^2}=5$.
Question 3 of 12
A mass $m$ on a spring has period $T$. If a mass $3m$ is used, the new period is:
$T$
$\sqrt{3}\,T$
$3T$
$\frac{T}{\sqrt{3}}$
Explanation: $T\propto\sqrt{m}$, so tripling mass multiplies the period by $\sqrt{3}$.
Question 4 of 12
A particle in SHM takes time $t$ to move from the mean position to half the amplitude. The period is (motion starts at the mean position):
$6t$
$12t$
$8t$
$4t$
Explanation: With $x=A\sin\omega t$, $\frac{A}{2}=A\sin\omega t\Rightarrow\omega t=\frac{\pi}{6}\Rightarrow t=\frac{T}{12}$, so $T=12t$.
Question 5 of 12
If the length of a seconds pendulum (period 2 s) is increased by 21%, the new period is approximately:
$2.0\ \text{s}$
$2.1\ \text{s}$
$2.2\ \text{s}$
$2.4\ \text{s}$
Explanation: $T\propto\sqrt{L}$; $\sqrt{1.21}=1.1$, so the new period is $2\times1.1=2.2\ \text{s}$.
Question 6 of 12
The KE of an SHM particle is three times its PE. Its displacement is:
$\frac{A}{2}$
$\frac{A}{\sqrt{2}}$
$\frac{A}{\sqrt{3}}$
$\frac{A}{4}$
Explanation: $\frac{KE}{PE}=\frac{A^2-x^2}{x^2}=3\Rightarrow A^2=4x^2\Rightarrow x=\frac{A}{2}$.
Question 7 of 12
Two SHMs of the same amplitude and frequency act along the same line with a phase difference of $\frac{\pi}{2}$. The amplitude of the resultant is:
$A$
$2A$
$\sqrt{2}\,A$
$0$
Explanation: For equal amplitudes with phase difference $\frac{\pi}{2}$, the resultant amplitude is $\sqrt{A^2+A^2}=\sqrt{2}A$.
Question 8 of 12
A block on a spring oscillates with amplitude $A$. At what displacement is its speed half the maximum speed?
$\frac{A}{2}$
$\frac{A}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}A$
$\frac{A}{4}$
Explanation: $v=\frac{v_{max}}{2}\Rightarrow\omega\sqrt{A^2-x^2}=\frac{1}{2}\omega A\Rightarrow A^2-x^2=\frac{A^2}{4}\Rightarrow x=\frac{\sqrt{3}}{2}A$.
Question 9 of 12
A simple pendulum is mounted in a lift accelerating upward with acceleration $a$. Its period becomes:
$2\pi\sqrt{\frac{L}{g}}$
$2\pi\sqrt{\frac{L}{g+a}}$
$2\pi\sqrt{\frac{L}{g-a}}$
$2\pi\sqrt{\frac{L}{a}}$
Explanation: Effective gravity rises to $g+a$, so $T=2\pi\sqrt{\frac{L}{g+a}}$ (shorter period).
Question 10 of 12
The acceleration–displacement ($a$ vs $x$) graph of a particle in SHM is:
a parabola
a straight line with positive slope
a straight line with negative slope through the origin
a circle
Explanation: $a=-\omega^2 x$ is linear in $x$ with negative slope $-\omega^2$ passing through the origin.
Question 11 of 12
A particle executes SHM with period $T$. The time taken to travel from $x=0$ to $x=\frac{A}{\sqrt{2}}$ (starting at the mean position) is:
$\frac{T}{4}$
$\frac{T}{8}$
$\frac{T}{6}$
$\frac{T}{12}$
Explanation: $\frac{A}{\sqrt{2}}=A\sin\omega t\Rightarrow\omega t=\frac{\pi}{4}\Rightarrow t=\frac{T}{8}$.
Question 12 of 12
When a spring is cut into two equal halves, the force constant of each half becomes:
$\frac{k}{2}$
$k$
$2k$
$4k$
Explanation: Force constant is inversely proportional to length, so halving the length doubles the constant to $2k$.