Energy in SHM — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Energy in SHM MCQs with step-by-step solutions (33 questions). Part of Oscillations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Energy in SHM · easy · theory
The total energy of a particle of mass $m$ performing S.H.M. of amplitude $A$ and angular frequency $\omega$ is:
A. $\dfrac{1}{2}m\omega A^2$
B. $\dfrac{1}{2}m\omega^2A$
C. $m\omega^2A^2$
D. $\dfrac{1}{2}m\omega^2A^2$ ✓ Correct
Solution: At an extreme position all the energy is potential: $E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}m\omega^2A^2$, since $k = m\omega^2$.
Q2 — Energy in SHM · easy · theory
The potential energy of a particle performing S.H.M. at displacement $x$ from the mean position is:
A. $\dfrac{1}{2}k(A^2 - x^2)$
B. $kx^2$
C. $\dfrac{1}{2}kA^2$
D. $\dfrac{1}{2}kx^2$ ✓ Correct
Solution: The restoring force is $F = -kx$, so the work stored in displacing the particle to $x$ is $U = \dfrac{1}{2}kx^2$.
Q3 — Energy in SHM · medium · theory
The kinetic energy of a particle performing S.H.M. at displacement $x$ is:
A. $\dfrac{1}{2}k(A^2 - x^2)$ ✓ Correct
B. $\dfrac{1}{2}kA^2$
C. $\dfrac{1}{2}kx^2$
D. $\dfrac{1}{2}k(A + x)^2$
Solution: Since the total energy $\dfrac{1}{2}kA^2$ is shared between kinetic and potential, $K = E - U = \dfrac{1}{2}k(A^2 - x^2)$.
Q4 — Energy in SHM · medium · numerical
A simple harmonic oscillator has total mechanical energy $E$. Its potential energy at displacement $x = \dfrac{A}{2}$ is:
A. $\dfrac{E}{4}$ ✓ Correct
B. $\dfrac{E}{2}$
C. $\dfrac{E}{\sqrt{2}}$
D. $\dfrac{3E}{4}$
Solution: $U = \dfrac{1}{2}kx^2 = \dfrac{1}{2}k\left(\dfrac{A}{2}\right)^2 = \dfrac{1}{4}\left(\dfrac{1}{2}kA^2\right) = \dfrac{E}{4}$.
Q5 — Energy in SHM · medium · numerical
At what displacement from the mean position are the kinetic and potential energies of a particle in S.H.M. equal?
A. $\dfrac{\sqrt{3}A}{2}$
B. $\dfrac{A}{2}$
C. $\dfrac{A}{\sqrt{3}}$
D. $\dfrac{A}{\sqrt{2}}$ ✓ Correct
Solution: Setting $\dfrac{1}{2}k(A^2 - x^2) = \dfrac{1}{2}kx^2$ gives $A^2 = 2x^2$, so $x = \dfrac{A}{\sqrt{2}}$.
Q6 — Energy in SHM · easy · theory
The total mechanical energy of a particle performing S.H.M. (in the absence of damping) is:
A. Constant throughout the motion ✓ Correct
B. Zero at the mean position
C. Maximum at the extreme positions only
D. Maximum at the mean position only
Solution: Kinetic and potential energies interchange continuously, but their sum $\dfrac{1}{2}kA^2$ stays fixed because the restoring force is conservative.
Q7 — Energy in SHM · easy · theory
The total energy of a particle performing S.H.M. is proportional to:
A. The inverse of the amplitude
B. The amplitude
C. The cube of the amplitude
D. The square of the amplitude ✓ Correct
Solution: $E = \dfrac{1}{2}kA^2$, so $E \propto A^2$.
Q8 — Energy in SHM · easy · numerical
If the amplitude of a simple harmonic oscillator is doubled without changing its frequency, its total energy becomes:
A. Half
B. Unchanged
C. Four times ✓ Correct
D. Twice
Solution: Since $E \propto A^2$, doubling the amplitude multiplies the energy by $4$.
Q9 — Energy in SHM · easy · theory
The kinetic energy of a particle performing S.H.M. is maximum at the:
A. Mean position ✓ Correct
B. Position $x = A/2$
C. Extreme positions
D. Position $x = A/\sqrt{2}$
Solution: The speed is greatest at $x = 0$, where the entire energy of the oscillator is kinetic.
Q10 — Energy in SHM · hard · theory
The frequency with which the kinetic energy of a simple harmonic oscillator varies is:
A. Twice the frequency of the oscillation ✓ Correct
B. Independent of the frequency of the oscillation
C. Equal to the frequency of the oscillation
D. Half the frequency of the oscillation
Solution: Kinetic energy depends on $v^2$, and $\cos^2\omega t = \dfrac{1 + \cos 2\omega t}{2}$. The energy therefore completes two maxima in each full oscillation.
Q11 — Energy in SHM · medium · numerical
A particle of mass $0.2\text{ kg}$ performs S.H.M. of amplitude $0.1\text{ m}$ with angular frequency $10\text{ rad/s}$. Its total energy is:
A. $0.05\text{ J}$
B. $0.2\text{ J}$
C. $1.0\text{ J}$
D. $0.1\text{ J}$ ✓ Correct
Solution: $E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{1}{2}(0.2)(100)(0.01) = 0.1\text{ J}$.
Q12 — Energy in SHM · medium · numerical
At displacement $x = \dfrac{A}{2}$, the ratio of the kinetic energy to the total energy of a particle in S.H.M. is:
A. $\dfrac{1}{4}$
B. $\dfrac{3}{4}$ ✓ Correct
C. $\dfrac{1}{3}$
D. $\dfrac{1}{2}$
Solution: $\dfrac{K}{E} = \dfrac{A^2 - x^2}{A^2} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.
Q13 — Energy in SHM · easy · theory
The graph of the potential energy of a simple harmonic oscillator against displacement is:
A. A sine curve
B. A circle
C. A parabola ✓ Correct
D. A straight line
Solution: Since $U = \dfrac{1}{2}kx^2$ is quadratic in $x$, the graph is a parabola with its minimum at the mean position.
Q14 — Energy in SHM · hard · theory
The average kinetic energy of a simple harmonic oscillator over one complete oscillation is:
A. Zero
B. $E$
C. $\dfrac{E}{4}$
D. $\dfrac{E}{2}$ ✓ Correct
Solution: Averaged over a full cycle, the kinetic and potential energies are equal, and each is half the constant total energy.
Q15 — Energy in SHM · medium · theory
The total energy of a simple harmonic oscillator is proportional to the square of its:
A. Time period
B. Displacement
C. Phase constant
D. Angular frequency ✓ Correct
Solution: From $E = \dfrac{1}{2}m\omega^2A^2$, the energy varies as $\omega^2$ for a given mass and amplitude.
Q16 — Energy in SHM · easy · theory
When a simple harmonic oscillator passes through its mean position, its potential energy is:
A. Equal to the total energy
B. Zero, and the kinetic energy equals the total energy ✓ Correct
C. Half the total energy
D. Maximum
Solution: At $x = 0$, $U = \dfrac{1}{2}kx^2 = 0$, so the whole of the constant total energy appears as kinetic energy.
Q17 — Energy in SHM · medium · numerical
Two simple harmonic oscillators have the same mass and amplitude, but the frequency of the second is twice that of the first. The ratio of their total energies $E_1 : E_2$ is:
A. $1 : 4$ ✓ Correct
B. $4 : 1$
C. $2 : 1$
D. $1 : 2$
Solution: With $m$ and $A$ common, $E \propto \omega^2$. Doubling the frequency quadruples the energy, so $E_1 : E_2 = 1 : 4$.
Q18 — Energy in SHM · medium · numerical
A particle of mass $0.5\text{ kg}$ performs S.H.M. of amplitude $0.02\text{ m}$ with angular frequency $20\text{ rad/s}$. Its total energy is:
A. $0.4\text{ J}$
B. $0.08\text{ J}$
C. $0.02\text{ J}$
D. $0.04\text{ J}$ ✓ Correct
Solution: $E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{1}{2}(0.5)(400)(0.0004) = 0.04\text{ J}$.
Q19 — Energy in SHM · medium · numerical
A spring of force constant $800\text{ N/m}$ oscillates with an amplitude of $0.05\text{ m}$. The total energy of the oscillation is:
A. $1\text{ J}$ ✓ Correct
B. $40\text{ J}$
C. $0.5\text{ J}$
D. $2\text{ J}$
Solution: $E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}(800)(0.0025) = 1\text{ J}$.
Q20 — Energy in SHM · hard · numerical
The total energy of a simple harmonic oscillator is $2\text{ J}$. Its kinetic energy at a displacement of half the amplitude is:
A. $1.0\text{ J}$
B. $2.0\text{ J}$
C. $0.5\text{ J}$
D. $1.5\text{ J}$ ✓ Correct
Solution: $\dfrac{K}{E} = 1 - \dfrac{x^2}{A^2} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$, so $K = 0.75 \times 2 = 1.5\text{ J}$.
Q21 — Energy in SHM · easy · numerical
If the amplitude of a simple harmonic oscillator is tripled at constant frequency, its total energy becomes:
A. $27$ times
B. $6$ times
C. $3$ times
D. $9$ times ✓ Correct
Solution: $E \propto A^2$, so tripling the amplitude multiplies the energy by $3^2 = 9$.
Q22 — Energy in SHM · medium · numerical
A particle of mass $1\text{ kg}$ performs S.H.M. of amplitude $0.05\text{ m}$ with angular frequency $10\text{ rad/s}$. Its total energy is:
A. $0.0625\text{ J}$
B. $0.25\text{ J}$
C. $0.125\text{ J}$ ✓ Correct
D. $1.25\text{ J}$
Solution: $E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{1}{2}(1)(100)(0.0025) = 0.125\text{ J}$.
Q23 — Energy in SHM · medium · numerical
A spring of force constant $200\text{ N/m}$ oscillates with an amplitude of $0.1\text{ m}$. The total energy is:
A. $20\text{ J}$
B. $2\text{ J}$
C. $1\text{ J}$ ✓ Correct
D. $0.5\text{ J}$
Solution: $E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}(200)(0.01) = 1\text{ J}$.
Q24 — Energy in SHM · medium · numerical
The total energy of a simple harmonic oscillator is $8\text{ J}$. Its kinetic energy at a displacement of half the amplitude is:
A. $4\text{ J}$
B. $2\text{ J}$
C. $6\text{ J}$ ✓ Correct
D. $8\text{ J}$
Solution: $\dfrac{K}{E} = 1 - \dfrac{x^2}{A^2} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$, so $K = 0.75 \times 8 = 6\text{ J}$.
Q25 — Energy in SHM · medium · numerical
The total energy of a simple harmonic oscillator is $8\text{ J}$. Its potential energy at a displacement of half the amplitude is:
A. $1\text{ J}$
B. $2\text{ J}$ ✓ Correct
C. $6\text{ J}$
D. $4\text{ J}$
Solution: $U = \dfrac{1}{2}kx^2 = \dfrac{1}{4}\left(\dfrac{1}{2}kA^2\right) = \dfrac{8}{4} = 2\text{ J}$.
Q26 — Energy in SHM · easy · numerical
If the amplitude of a simple harmonic oscillator is doubled at constant frequency, its total energy becomes:
A. Unchanged
B. $8$ times
C. $2$ times
D. $4$ times ✓ Correct
Solution: $E \propto A^2$, so doubling the amplitude quadruples the energy.
Q27 — Energy in SHM · easy · numerical
If the angular frequency of a simple harmonic oscillator is doubled at constant amplitude, its total energy becomes:
A. $4$ times ✓ Correct
B. $8$ times
C. Half
D. $2$ times
Solution: $E = \dfrac{1}{2}m\omega^2A^2 \propto \omega^2$, so doubling $\omega$ quadruples the energy.
Q28 — Energy in SHM · medium · numerical
The total energy of a simple harmonic oscillator is $10\text{ J}$. Its kinetic energy at a displacement of $\dfrac{A}{\sqrt{2}}$ is:
A. $5\text{ J}$ ✓ Correct
B. $2.5\text{ J}$
C. $7.5\text{ J}$
D. $10\text{ J}$
Solution: At $x = \dfrac{A}{\sqrt{2}}$, $\dfrac{x^2}{A^2} = \dfrac{1}{2}$, so kinetic and potential energies are equal at $\dfrac{E}{2} = 5\text{ J}$.
Q29 — Energy in SHM · hard · numerical
A particle of mass $0.1\text{ kg}$ performs S.H.M. of amplitude $0.03\text{ m}$ with angular frequency $20\text{ rad/s}$. Its total energy is:
A. $0.09\text{ J}$
B. $0.018\text{ J}$ ✓ Correct
C. $0.0009\text{ J}$
D. $0.036\text{ J}$
Solution: $E = \dfrac{1}{2}(0.1)(400)(0.0009) = 0.018\text{ J}$.
Q30 — Energy in SHM · medium · numerical
A spring of force constant $400\text{ N/m}$ oscillates with an amplitude of $0.2\text{ m}$. The total energy is:
A. $40\text{ J}$
B. $16\text{ J}$
C. $8\text{ J}$ ✓ Correct
D. $4\text{ J}$
Solution: $E = \dfrac{1}{2}(400)(0.04) = 8\text{ J}$.