Class 11 Physics Chapter 13: Oscillations worksheet with answer key
A 20-mark worksheet on Oscillations for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (d) stays the same
T = 2π√(m/k) contains neither the amplitude nor the energy; the period of SHM does not depend on the amplitude.
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Answer: (a) 10%
T ∝ √L, so T′/T = √1.21 = 1.1: an increase of 10%. 21% treats T as proportional to L; 42% doubles the change instead of halving it.
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Answer: (d) 31.4 rad s⁻¹
ω = 2π/T = 2π/0.2 = 10π ≈ 31.4 rad s⁻¹. 5 rad s⁻¹ is the frequency 1/T, not ω; 15.7 rad s⁻¹ uses π/T; 1.26 rad s⁻¹ multiplies 2π by T.
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Answer: (a) π
With x = A cos ωt, a = −ω²A cos ωt = ω²A cos(ωt + π); the acceleration plot differs in phase from the displacement plot by π. π/2 is the phase difference between displacement and velocity.
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Answer: (b) becomes four times
E = ½kA². With k unchanged, doubling A multiplies E by 2² = 4.
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ν = ω/2π = 1000π/2π = 500 Hz. T = 1/ν = 0.002 s, so 50 vibrations take 50 × 0.002 = 0.1 s.
- ν = ω/2π = 500 Hz (1)
- T = 0.002 s; 50 vibrations in 0.1 s (1)
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The shadow is the projection of the uniform circular motion on a diameter of the circle, so it executes simple harmonic motion along a horizontal line, about the point opposite the centre. Its amplitude is the radius, 30 cm, and its period is the time for one revolution, 1/2 s = 0.5 s.
- projection of uniform circular motion: SHM (1)
- amplitude 30 cm, period 0.5 s (1)
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The swinging dies out because of damping: friction at the support and air resistance take energy out of the motion, until the swing comes to rest at its equilibrium position. It can be kept going by an external periodic agency, for example a push given once in every swing, which supplies the energy lost.
- damping by friction and air resistance removes energy; comes to rest at equilibrium (1)
- an external periodic push maintains the oscillation (1)
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Two forces act: the tension T along the string towards the support, and the weight mg vertically down, which resolves into mg cos θ along the string (away from the support) and mg sin θ perpendicular to it. T − mg cos θ provides the radial (centripetal) acceleration and has no torque about the support; mg sin θ gives the tangential acceleration and the restoring torque. T = mg cos θ + mv²/L is greatest at the mean position, where cos θ = 1 and the speed is greatest.
- tension and weight; mg cos θ along, mg sin θ perpendicular (1)
- T − mg cos θ radial; mg sin θ tangential, restoring torque (1)
- tension greatest at the mean position: cos θ largest and speed largest (1)
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ω = √(k/m) = √(200/0.5) = 20 rad s⁻¹. (a) vm = ωA gives A = 2/20 = 0.1 m. (b) E = ½mvm² = ½ × 0.5 × 4 = 1.0 J (check: ½kA² = ½ × 200 × 0.01 = 1 J). (c) v = ω√(A² − x²): 1 = 20√(0.01 − x²), so x² = 0.01 − 0.0025 = 0.0075 and x ≈ 0.0866 m, about 8.7 cm.
- ω = 20 rad s⁻¹, A = 0.1 m (1)
- E = 1.0 J (1)
- x ≈ 0.0866 m (1)
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v = dx/dt = −ωA sin(ωt + φ) = ωA cos(ωt + φ + π/2). a = dv/dt = −ω²A cos(ωt + φ) = −ω²x. So the velocity leads the displacement in phase by π/2, with amplitude ωA, and the acceleration differs from it in phase by π, with amplitude ω²A.
- v = −ωA sin(ωt + φ) (1)
- a = −ω²A cos(ωt + φ) = −ω²x (1)
- phase differences π/2 and π (1)