Question

for a simple harmonic oscillator of mass M =10kgs and amplitude A=2 m the variation of potential energy u as a function of displacement X from mean position is as shown the time period of oscillation in second is​

Answer 1

Explanation:

For a simple harmonic oscillator of mass m = 10 kg and amplitude A = 2 m the variation of potential energy U as a function of displacement x from mean position is as shown. The time period of oscillation in second, is U 30 J 10 J 0 -A (1) A (2) 21 3/17 (3) Na (4) 4 It has been observed that velocity of longitudinal

Answer 2

Disclaimer:

One image of the variation of the potential energy graph is attached.

Given:

Mass, m = 10 kg

Amplitude, A = 2 m

Find:

The time period of the oscillation.

Solution:

Kinetic energy at the mean position can be calculated as the difference between total energy and potential energy at the mean position.

From the graph,

KE = TE − PE = 30 - 10

KE = 20  J

Kinetic energy at the mean position, KE = 1/2 mv²

(1/2) mv² = 20

(1/2) ×10 v² = 20

v² = 4

v = √4 = 2 m/s

According to a simple harmonic oscillator,

v = Aω and ω = 2π/T

v = A×2π/T

T = A×2π/v = 2×2π/2

T = 2π s

Hence, the time period of the oscillation is 2π seconds.

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