Superposition of Oscillations — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Superposition of Oscillations 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 — Superposition of Oscillations · medium · theory
Two S.H.M.s of amplitudes $A_1$ and $A_2$ along the same line with a phase difference $\phi$ combine to give a resultant amplitude of:
A. $A_1 + A_2$
B. $\sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}$ ✓ Correct
C. $\sqrt{A_1^2 + A_2^2 - 2A_1A_2\cos\phi}$
D. $\sqrt{A_1^2 + A_2^2}$
Solution: The two motions add like vectors (phasors) inclined at the phase angle $\phi$, giving the usual parallelogram result.
Q2 — Superposition of Oscillations · medium · numerical
Two simple harmonic motions are represented by $x_1 = 4\sin(\omega t)$ and $x_2 = 3\cos(\omega t)$. The amplitude of the resultant motion is:
A. $1$
B. $5$ ✓ Correct
C. $12$
D. $7$
Solution: Writing $x_2 = 3\sin\left(\omega t + \dfrac{\pi}{2}\right)$ shows the phase difference is $90^\circ$, so $A = \sqrt{4^2 + 3^2} = 5$.
Q3 — Superposition of Oscillations · easy · theory
Two S.H.M.s of the same frequency along the same line are exactly in phase. The resultant amplitude is:
A. $A_1 + A_2$ ✓ Correct
B. $\sqrt{A_1^2 + A_2^2}$
C. Zero
D. $|A_1 - A_2|$
Solution: With $\phi = 0$, $\cos\phi = 1$ and the formula reduces to $\sqrt{(A_1 + A_2)^2} = A_1 + A_2$ — constructive superposition.
Q4 — Superposition of Oscillations · easy · theory
Two S.H.M.s of the same frequency along the same line differ in phase by $\pi$. The resultant amplitude is:
A. $|A_1 - A_2|$ ✓ Correct
B. $A_1 + A_2$
C. $A_1 A_2$
D. $\sqrt{A_1^2 + A_2^2}$
Solution: With $\cos\pi = -1$, the expression becomes $\sqrt{(A_1 - A_2)^2} = |A_1 - A_2|$ — destructive superposition.
Q5 — Superposition of Oscillations · hard · numerical
Two S.H.M.s of amplitudes $3\text{ cm}$ and $4\text{ cm}$ along the same line differ in phase by $60^\circ$. The resultant amplitude is approximately:
A. $7.00\text{ cm}$
B. $1.00\text{ cm}$
C. $6.08\text{ cm}$ ✓ Correct
D. $5.00\text{ cm}$
Solution: $A = \sqrt{9 + 16 + 2(3)(4)\cos 60^\circ} = \sqrt{25 + 12} = \sqrt{37} \approx 6.08\text{ cm}$.
Q6 — Superposition of Oscillations · hard · numerical
Two S.H.M.s of equal amplitude $A$ along the same line differ in phase by $\dfrac{\pi}{3}$. The resultant amplitude is:
A. $A$
B. $A\sqrt{2}$
C. $A\sqrt{3}$ ✓ Correct
D. $2A$
Solution: $A_R = \sqrt{A^2 + A^2 + 2A^2\cos 60^\circ} = \sqrt{2A^2 + A^2} = A\sqrt{3}$.
Q7 — Superposition of Oscillations · medium · theory
The superposition of two S.H.M.s of the same frequency along the same straight line produces:
A. Another S.H.M. of the same frequency ✓ Correct
B. A motion of double the frequency
C. A non-periodic motion
D. Uniform circular motion
Solution: Adding two sinusoids of identical frequency gives a single sinusoid of that frequency, with a new amplitude and phase.
Q8 — Superposition of Oscillations · hard · theory
Two S.H.M.s of equal amplitude and equal frequency acting along mutually perpendicular directions with a phase difference of $\dfrac{\pi}{2}$ produce a resultant path that is:
A. A parabola
B. A straight line
C. A circle ✓ Correct
D. An ellipse with unequal axes
Solution: With $x = A\sin\omega t$ and $y = A\cos\omega t$, squaring and adding gives $x^2 + y^2 = A^2$ — a circle.
Q9 — Superposition of Oscillations · hard · theory
Two S.H.M.s of the same frequency acting along mutually perpendicular directions and exactly in phase produce a resultant path that is:
A. A parabola
B. A circle
C. An ellipse
D. A straight line ✓ Correct
Solution: With $x = A_1\sin\omega t$ and $y = A_2\sin\omega t$, the ratio $\dfrac{y}{x} = \dfrac{A_2}{A_1}$ is constant, so the point moves along a straight line through the origin.
Q10 — Superposition of Oscillations · medium · theory
The closed curves traced by a point subjected to two perpendicular S.H.M.s of commensurable frequencies are known as:
A. Hysteresis loops
B. Lissajous figures ✓ Correct
C. Indicator diagrams
D. Fringe patterns
Solution: The shape of a Lissajous figure depends on the ratio of the two frequencies and on their phase difference, which makes it a useful frequency-comparison tool.
Q11 — Superposition of Oscillations · medium · theory
The phenomenon of beats arises from the superposition of two waves having:
A. Slightly different frequencies travelling in the same direction ✓ Correct
B. Exactly equal frequencies travelling in opposite directions
C. Mutually perpendicular directions of vibration
D. Very different frequencies
Solution: The resultant amplitude rises and falls at the difference frequency, which is heard as a periodic waxing and waning of loudness.
Q12 — Superposition of Oscillations · easy · theory
The principle of superposition of oscillations states that the resultant displacement at any instant is:
A. Always zero
B. The vector sum of the individual displacements ✓ Correct
C. The product of the individual displacements
D. Always the larger of the two displacements
Solution: Because the governing equation is linear, each oscillation proceeds as if the other were absent, and their displacements simply add.
Q13 — Superposition of Oscillations · hard · numerical
Two S.H.M.s of amplitudes $5\text{ cm}$ and $5\text{ cm}$ along the same line differ in phase by $120^\circ$. The resultant amplitude is:
A. $5\text{ cm}$ ✓ Correct
B. $10\text{ cm}$
C. $8.66\text{ cm}$
D. Zero
Solution: $A_R = \sqrt{25 + 25 + 2(25)\cos 120^\circ} = \sqrt{50 - 25} = \sqrt{25} = 5\text{ cm}$.
Q14 — Superposition of Oscillations · easy · theory
The resultant amplitude of two superposed S.H.M.s along the same line is maximum when the phase difference between them is:
A. $\dfrac{3\pi}{2}$
B. $\pi$
C. Zero ✓ Correct
D. $\dfrac{\pi}{2}$
Solution: The term $2A_1A_2\cos\phi$ is largest when $\cos\phi = 1$, i.e. when the two motions are in step.
Q15 — Superposition of Oscillations · medium · theory
Two S.H.M.s along the same line have equal amplitudes and a phase difference of $\pi$. The resultant motion is:
A. An S.H.M. of double the amplitude
B. An S.H.M. of double the frequency
C. Circular motion
D. Complete rest, since the two exactly cancel ✓ Correct
Solution: With $A_1 = A_2$ and $\phi = \pi$, the resultant amplitude $|A_1 - A_2|$ is zero, so the particle stays at rest.
Q16 — Superposition of Oscillations · hard · theory
Two perpendicular S.H.M.s of unequal amplitudes with a phase difference of $\dfrac{\pi}{2}$ trace a path that is:
A. A circle
B. An ellipse with axes along the two directions ✓ Correct
C. A straight line
D. A parabola
Solution: With $x = A_1\sin\omega t$ and $y = A_2\cos\omega t$, eliminating $t$ gives $\dfrac{x^2}{A_1^2} + \dfrac{y^2}{A_2^2} = 1$ — the equation of an ellipse.
Q17 — Superposition of Oscillations · medium · theory
Superposition of two S.H.M.s is governed by a simple additive rule because the equation of simple harmonic motion is:
A. Quadratic in the displacement
B. Linear in the displacement ✓ Correct
C. Independent of the displacement
D. Non-linear in the velocity
Solution: For a linear differential equation, any sum of solutions is itself a solution — which is precisely what the superposition principle asserts.
Q18 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitudes $6\text{ cm}$ and $8\text{ cm}$ along the same line differ in phase by $90^\circ$. The resultant amplitude is:
A. $2\text{ cm}$
B. $10\text{ cm}$ ✓ Correct
C. $7\text{ cm}$
D. $14\text{ cm}$
Solution: With $\cos 90^\circ = 0$, $A = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}$.
Q19 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitude $4\text{ cm}$ each along the same line differ in phase by $180^\circ$. The resultant amplitude is:
A. $5.66\text{ cm}$
B. $4\text{ cm}$
C. $8\text{ cm}$
D. Zero ✓ Correct
Solution: With $\cos 180^\circ = -1$, $A = \sqrt{16 + 16 - 32} = 0$ — the two motions cancel exactly.
Q20 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitudes $5\text{ cm}$ and $12\text{ cm}$ along the same line differ in phase by $90^\circ$. The resultant amplitude is:
A. $8.5\text{ cm}$
B. $17\text{ cm}$
C. $13\text{ cm}$ ✓ Correct
D. $7\text{ cm}$
Solution: $A = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$.
Q21 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitude $3\text{ cm}$ each along the same line differ in phase by $90^\circ$. The resultant amplitude is approximately:
A. $3.00\text{ cm}$
B. $5.20\text{ cm}$
C. $6.00\text{ cm}$
D. $4.24\text{ cm}$ ✓ Correct
Solution: $A = \sqrt{9 + 9} = 3\sqrt{2} \approx 4.24\text{ cm}$.
Q22 — Superposition of Oscillations · easy · numerical
Two S.H.M.s of amplitudes $8\text{ cm}$ and $6\text{ cm}$ along the same line are exactly in phase. The resultant amplitude is:
A. $10\text{ cm}$
B. $2\text{ cm}$
C. $48\text{ cm}$
D. $14\text{ cm}$ ✓ Correct
Solution: With $\phi = 0$ the amplitudes add directly: $A = 8 + 6 = 14\text{ cm}$.
Q23 — Superposition of Oscillations · easy · numerical
Two S.H.M.s of amplitudes $8\text{ cm}$ and $6\text{ cm}$ along the same line differ in phase by $180^\circ$. The resultant amplitude is:
A. $14\text{ cm}$
B. Zero
C. $2\text{ cm}$ ✓ Correct
D. $10\text{ cm}$
Solution: With $\phi = \pi$ the amplitudes subtract: $A = |8 - 6| = 2\text{ cm}$.
Q24 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitudes $9\text{ cm}$ and $12\text{ cm}$ along the same line differ in phase by $90^\circ$. The resultant amplitude is:
A. $15\text{ cm}$ ✓ Correct
B. $10.5\text{ cm}$
C. $3\text{ cm}$
D. $21\text{ cm}$
Solution: $A = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}$.
Q25 — Superposition of Oscillations · hard · numerical
Two S.H.M.s of amplitude $5\text{ cm}$ each along the same line differ in phase by $60^\circ$. The resultant amplitude is approximately:
A. $8.66\text{ cm}$ ✓ Correct
B. $10.0\text{ cm}$
C. $5.0\text{ cm}$
D. $7.07\text{ cm}$
Solution: $A = \sqrt{25 + 25 + 2(25)\cos 60^\circ} = \sqrt{50 + 25} = \sqrt{75} \approx 8.66\text{ cm}$.
Q26 — Superposition of Oscillations · hard · numerical
Two S.H.M.s of amplitude $6\text{ cm}$ each along the same line differ in phase by $120^\circ$. The resultant amplitude is:
A. $12\text{ cm}$
B. $6\text{ cm}$ ✓ Correct
C. Zero
D. $10.4\text{ cm}$
Solution: $A = \sqrt{36 + 36 + 2(36)\cos 120^\circ} = \sqrt{72 - 36} = \sqrt{36} = 6\text{ cm}$.
Q27 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitudes $7\text{ cm}$ and $24\text{ cm}$ along the same line differ in phase by $90^\circ$. The resultant amplitude is:
A. $17\text{ cm}$
B. $31\text{ cm}$
C. $21\text{ cm}$
D. $25\text{ cm}$ ✓ Correct
Solution: $A = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$.
Q28 — Superposition of Oscillations · hard · numerical
Two S.H.M.s of amplitudes $2\text{ cm}$ and $3\text{ cm}$ along the same line differ in phase by $60^\circ$. The resultant amplitude is approximately:
A. $5.00\text{ cm}$
B. $3.61\text{ cm}$
C. $4.36\text{ cm}$ ✓ Correct
D. $1.00\text{ cm}$
Solution: $A = \sqrt{4 + 9 + 2(2)(3)(0.5)} = \sqrt{13 + 6} = \sqrt{19} \approx 4.36\text{ cm}$.
Q29 — Superposition of Oscillations · easy · numerical
Two S.H.M.s of amplitude $3\text{ cm}$ each along the same line are in phase. The resultant amplitude is:
A. $3\text{ cm}$
B. $6\text{ cm}$ ✓ Correct
C. Zero
D. $4.24\text{ cm}$
Solution: In-phase superposition gives $A = 3 + 3 = 6\text{ cm}$.
Q30 — Superposition of Oscillations · medium · numerical
Two S.H.M.s of amplitude $5\text{ cm}$ each along the same line differ in phase by $90^\circ$. The resultant amplitude is approximately:
A. $5.0\text{ cm}$
B. $7.07\text{ cm}$ ✓ Correct
C. $8.66\text{ cm}$
D. $10.0\text{ cm}$
Solution: $A = \sqrt{25 + 25} = 5\sqrt{2} \approx 7.07\text{ cm}$.