Simple Harmonic Motion Basics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Simple Harmonic Motion Basics MCQs with step-by-step solutions (34 questions). Part of Oscillations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Simple Harmonic Motion Basics · easy · theory
A motion is simple harmonic if the acceleration of the particle is:
A. Proportional to the velocity of the particle
B. Proportional to the displacement and directed towards the mean position ✓ Correct
C. Proportional to the displacement and directed away from the mean position
D. Constant in magnitude and direction
Solution: The defining relation is $a = -\omega^2 x$. The minus sign shows the acceleration is always a restoring one, pointing back towards the equilibrium position.
Q2 — Simple Harmonic Motion Basics · easy · numerical
A particle executes S.H.M. described by $x = 0.05\sin(100t + \pi/6)\text{ m}$. Its maximum velocity is:
A. $0.5\text{ m/s}$
B. $2.5\text{ m/s}$
C. $50\text{ m/s}$
D. $5\text{ m/s}$ ✓ Correct
Solution: Here $A = 0.05\text{ m}$ and $\omega = 100\text{ rad/s}$, so $v_{\max} = A\omega = 0.05 \times 100 = 5\text{ m/s}$.
Q3 — Simple Harmonic Motion Basics · easy · theory
The maximum acceleration of a particle performing S.H.M. of amplitude $A$ and angular frequency $\omega$ is:
A. $A\omega^2$ ✓ Correct
B. $A^2\omega$
C. $A\omega$
D. $\dfrac{A}{\omega^2}$
Solution: Since $a = -\omega^2 x$, the magnitude is greatest at the extreme positions where $|x| = A$, giving $a_{\max} = A\omega^2$.
Q4 — Simple Harmonic Motion Basics · medium · theory
The velocity of a particle performing S.H.M. of amplitude $A$ at displacement $x$ from the mean position is:
A. $\omega\sqrt{A^2 + x^2}$
B. $\omega\sqrt{A^2 - x^2}$ ✓ Correct
C. $\omega(A - x)$
D. $\omega^2\sqrt{A^2 - x^2}$
Solution: From energy conservation, $\dfrac{1}{2}mv^2 = \dfrac{1}{2}m\omega^2(A^2 - x^2)$, giving $v = \omega\sqrt{A^2 - x^2}$.
Q5 — Simple Harmonic Motion Basics · hard · numerical
A particle in linear S.H.M. has velocities $4\text{ m/s}$ and $3\text{ m/s}$ at displacements $3\text{ cm}$ and $4\text{ cm}$ respectively. The amplitude of oscillation is:
A. $2.5\text{ cm}$
B. $1\text{ cm}$
C. $5\text{ cm}$ ✓ Correct
D. $7\text{ cm}$
Solution: Using $v^2 = \omega^2(A^2 - x^2)$ twice and dividing: $\dfrac{16}{9} = \dfrac{A^2 - 9}{A^2 - 16}$. Cross-multiplying gives $7A^2 = 175$, so $A^2 = 25$ and $A = 5\text{ cm}$.
Q6 — Simple Harmonic Motion Basics · hard · numerical
A particle executes linear S.H.M. of period $8\text{ s}$ and amplitude $10\text{ cm}$. The time it takes to move from the mean position to a displacement of $5\text{ cm}$ is:
A. $\dfrac{1}{3}\text{ s}$
B. $2\text{ s}$
C. $\dfrac{2}{3}\text{ s}$ ✓ Correct
D. $1\text{ s}$
Solution: Starting at the mean position, $x = A\sin\omega t$. Setting $x = \dfrac{A}{2}$ gives $\sin\omega t = 0.5$, so $\omega t = \dfrac{\pi}{6}$ and $t = \dfrac{T}{12} = \dfrac{8}{12} = \dfrac{2}{3}\text{ s}$.
Q7 — Simple Harmonic Motion Basics · easy · theory
At the mean position of a particle performing S.H.M.:
A. Both velocity and acceleration are maximum
B. The velocity is zero and the acceleration is maximum
C. The velocity is maximum and the acceleration is zero ✓ Correct
D. Both velocity and acceleration are zero
Solution: At $x = 0$ the restoring force vanishes so $a = 0$, while all the energy is kinetic so the speed is at its greatest.
Q8 — Simple Harmonic Motion Basics · easy · theory
At the extreme positions of a particle performing S.H.M.:
A. The velocity is maximum and the acceleration is zero
B. The velocity is zero and the acceleration is maximum ✓ Correct
C. Both are maximum
D. Both are zero
Solution: At $|x| = A$ the particle momentarily stops before reversing, and the restoring force — and hence the acceleration — is at its largest.
Q9 — Simple Harmonic Motion Basics · easy · theory
The period of oscillation of a mass $m$ suspended from a spring of force constant $k$ is:
A. $2\pi\dfrac{m}{k}$
B. $2\pi\sqrt{\dfrac{k}{m}}$
C. $2\pi\sqrt{\dfrac{m}{k}}$ ✓ Correct
D. $\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}$
Solution: With $\omega = \sqrt{\dfrac{k}{m}}$, the period is $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{m}{k}}$.
Q10 — Simple Harmonic Motion Basics · medium · numerical
A mass $m$ is suspended from two identical vertical springs, each of force constant $k$, connected in parallel. The period of vertical oscillation is:
A. $\pi\sqrt{\dfrac{m}{k}}$
B. $2\pi\sqrt{\dfrac{2m}{k}}$
C. $2\pi\sqrt{\dfrac{m}{k}}$
D. $2\pi\sqrt{\dfrac{m}{2k}}$ ✓ Correct
Solution: Springs in parallel share the extension, so their constants add: $k_{eq} = 2k$. Hence $T = 2\pi\sqrt{\dfrac{m}{2k}}$.
Q11 — Simple Harmonic Motion Basics · medium · theory
Two identical springs, each of force constant $k$, are joined in series. The equivalent force constant of the combination is:
A. $k$
B. $4k$
C. $\dfrac{k}{2}$ ✓ Correct
D. $2k$
Solution: For springs in series the extensions add, so $\dfrac{1}{k_{eq}} = \dfrac{1}{k} + \dfrac{1}{k} = \dfrac{2}{k}$, giving $k_{eq} = \dfrac{k}{2}$.
Q12 — Simple Harmonic Motion Basics · medium · theory
A uniform spring of force constant $k$ is cut into two equal halves. The force constant of each half is:
A. $4k$
B. $k$
C. $2k$ ✓ Correct
D. $\dfrac{k}{2}$
Solution: The force constant is inversely proportional to the free length, since a shorter spring stretches less for the same load. Halving the length therefore doubles $k$.
Q13 — Simple Harmonic Motion Basics · medium · theory
Simple harmonic motion can be regarded geometrically as:
A. The projection of uniform circular motion on a diameter ✓ Correct
B. Motion along a parabola
C. Uniformly accelerated motion along a line
D. The projection of uniform circular motion on the circumference
Solution: If a point moves uniformly on a circle of radius $A$, its shadow on any diameter executes S.H.M. of amplitude $A$ and the same angular frequency.
Q14 — Simple Harmonic Motion Basics · easy · theory
The quantity $(\omega t + \phi)$ in the S.H.M. equation $x = A\sin(\omega t + \phi)$ is called the:
A. Frequency of the motion
B. Amplitude of the motion
C. Time period
D. Phase of the motion ✓ Correct
Solution: The phase specifies the state of the oscillator at a given instant, and $\phi$, its value at $t = 0$, is the initial phase or epoch.
Q15 — Simple Harmonic Motion Basics · easy · theory
A particle performs S.H.M. with angular frequency $\omega$. The period of its oscillation is:
A. $\dfrac{\omega}{2\pi}$
B. $\dfrac{\pi}{\omega}$
C. $\dfrac{2\pi}{\omega}$ ✓ Correct
D. $2\pi\omega$
Solution: One complete oscillation corresponds to a phase change of $2\pi$, so $T = \dfrac{2\pi}{\omega}$.
Q16 — Simple Harmonic Motion Basics · easy · numerical
If the mass attached to a spring is increased four times, the period of oscillation becomes:
A. Twice as large ✓ Correct
B. Unchanged
C. Four times as large
D. Half as large
Solution: $T = 2\pi\sqrt{\dfrac{m}{k}} \propto \sqrt{m}$, so quadrupling the mass doubles the period.
Q17 — Simple Harmonic Motion Basics · medium · theory
In S.H.M., the displacement and the acceleration of the particle differ in phase by:
A. $\dfrac{\pi}{4}\text{ rad}$
B. $\pi\text{ rad}$ ✓ Correct
C. Zero
D. $\dfrac{\pi}{2}\text{ rad}$
Solution: Since $a = -\omega^2 x$, the acceleration is always exactly opposite in sign to the displacement, which is a phase difference of $\pi$ radian.
Q18 — Simple Harmonic Motion Basics · easy · numerical
A particle executes S.H.M. given by $x = 0.1\sin(50t)\text{ m}$. Its maximum speed is:
A. $0.5\text{ m/s}$
B. $5\text{ m/s}$ ✓ Correct
C. $250\text{ m/s}$
D. $50\text{ m/s}$
Solution: $v_{\max} = A\omega = 0.1 \times 50 = 5\text{ m/s}$.
Q19 — Simple Harmonic Motion Basics · medium · numerical
For the S.H.M. $x = 0.1\sin(50t)\text{ m}$, the maximum acceleration is:
A. $250\text{ m/s}^2$ ✓ Correct
B. $500\text{ m/s}^2$
C. $25\text{ m/s}^2$
D. $5\text{ m/s}^2$
Solution: $a_{\max} = A\omega^2 = 0.1 \times 2500 = 250\text{ m/s}^2$.
Q20 — Simple Harmonic Motion Basics · medium · numerical
A particle performs S.H.M. with angular frequency $50\text{ rad/s}$. Its period is approximately:
A. $0.318\text{ s}$
B. $7.96\text{ s}$
C. $0.126\text{ s}$ ✓ Correct
D. $0.02\text{ s}$
Solution: $T = \dfrac{2\pi}{\omega} = \dfrac{6.283}{50} \approx 0.126\text{ s}$.
Q21 — Simple Harmonic Motion Basics · medium · numerical
A mass of $2\text{ kg}$ hangs from a spring of force constant $200\text{ N/m}$. Its period of oscillation is:
A. $0.314\text{ s}$
B. $0.1\text{ s}$
C. $1.256\text{ s}$
D. $0.628\text{ s}$ ✓ Correct
Solution: $T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{2}{200}} = 2\pi(0.1) \approx 0.628\text{ s}$.
Q22 — Simple Harmonic Motion Basics · hard · numerical
A particle performs S.H.M. of amplitude $5\text{ cm}$ with angular frequency $4\text{ rad/s}$. Its speed at a displacement of $3\text{ cm}$ is:
A. $8\text{ cm/s}$
B. $16\text{ cm/s}$ ✓ Correct
C. $20\text{ cm/s}$
D. $12\text{ cm/s}$
Solution: $v = \omega\sqrt{A^2 - x^2} = 4\sqrt{25 - 9} = 4 \times 4 = 16\text{ cm/s}$.
Q23 — Simple Harmonic Motion Basics · easy · numerical
A particle executes S.H.M. given by $x = 0.2\sin(10t)\text{ m}$. Its maximum speed is:
A. $20\text{ m/s}$
B. $10\text{ m/s}$
C. $2\text{ m/s}$ ✓ Correct
D. $0.2\text{ m/s}$
Solution: $v_{\max} = A\omega = 0.2 \times 10 = 2\text{ m/s}$.
Q24 — Simple Harmonic Motion Basics · medium · numerical
For the S.H.M. $x = 0.2\sin(10t)\text{ m}$, the maximum acceleration is:
A. $2\text{ m/s}^2$
B. $100\text{ m/s}^2$
C. $200\text{ m/s}^2$
D. $20\text{ m/s}^2$ ✓ Correct
Solution: $a_{\max} = A\omega^2 = 0.2 \times 100 = 20\text{ m/s}^2$.
Q25 — Simple Harmonic Motion Basics · easy · numerical
A particle performs S.H.M. with angular frequency $10\text{ rad/s}$. Its period is approximately:
A. $0.628\text{ s}$ ✓ Correct
B. $6.28\text{ s}$
C. $0.1\text{ s}$
D. $1.59\text{ s}$
Solution: $T = \dfrac{2\pi}{\omega} = \dfrac{6.283}{10} \approx 0.628\text{ s}$.
Q26 — Simple Harmonic Motion Basics · medium · numerical
A mass of $4\text{ kg}$ hangs from a spring of force constant $100\text{ N/m}$. Its period of oscillation is approximately:
A. $0.25\text{ s}$
B. $1.26\text{ s}$ ✓ Correct
C. $0.63\text{ s}$
D. $2.51\text{ s}$
Solution: $T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{4}{100}} = 2\pi(0.2) \approx 1.26\text{ s}$.
Q27 — Simple Harmonic Motion Basics · hard · numerical
A particle performs S.H.M. of amplitude $10\text{ cm}$ with angular frequency $5\text{ rad/s}$. Its speed at a displacement of $6\text{ cm}$ is:
A. $30\text{ cm/s}$
B. $80\text{ cm/s}$
C. $50\text{ cm/s}$
D. $40\text{ cm/s}$ ✓ Correct
Solution: $v = \omega\sqrt{A^2 - x^2} = 5\sqrt{100 - 36} = 5 \times 8 = 40\text{ cm/s}$.
Q28 — Simple Harmonic Motion Basics · hard · numerical
A particle performs S.H.M. of amplitude $13\text{ cm}$ with angular frequency $2\text{ rad/s}$. Its speed at a displacement of $5\text{ cm}$ is:
A. $16\text{ cm/s}$
B. $24\text{ cm/s}$ ✓ Correct
C. $10\text{ cm/s}$
D. $26\text{ cm/s}$
Solution: $v = 2\sqrt{169 - 25} = 2\sqrt{144} = 2 \times 12 = 24\text{ cm/s}$.
Q29 — Simple Harmonic Motion Basics · easy · numerical
A particle performs S.H.M. with a period of $4\text{ s}$. Its angular frequency is:
A. $25.1\text{ rad/s}$
B. $0.25\text{ rad/s}$
C. $1.57\text{ rad/s}$ ✓ Correct
D. $4\text{ rad/s}$
Solution: $\omega = \dfrac{2\pi}{T} = \dfrac{6.283}{4} \approx 1.57\text{ rad/s}$.
Q30 — Simple Harmonic Motion Basics · easy · numerical
A mass of $0.5\text{ kg}$ on a spring of force constant $50\text{ N/m}$ oscillates with angular frequency:
A. $10\text{ rad/s}$ ✓ Correct
B. $5\text{ rad/s}$
C. $25\text{ rad/s}$
D. $100\text{ rad/s}$
Solution: $\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{50}{0.5}} = \sqrt{100} = 10\text{ rad/s}$.