(a) Explanation Given data: Frequency of the simple harmonic motion is 3 Hz. Amplitude of the simple harmonic motion is 5 .08 × 10 − 2 m . Formula used: Write the expression for kinetic energy. K E = 1 2 m v 2 (1) Here, KE is the kinetic energy, m is the mass, and v is the velocity. Write the expression for the elastic potential energy. P E = 1 2 k x max 2 (2) Here, k is the spring constant and x max is the maximum compression of the spring. Write the equation for frequency. f = 1 2 π k m (3) Here, f is the frequency, k is the spring constant, and m is the mass. Explanation: Value of the maximum kinetic energy is, Substitute v max (initial velocity) for v in equation (1), K E = 1 2 m v max 2 (4) The amplitude of the spring is the maximum compression of the spring, that is, the maximum displacement of the simple harmonic wave from the reference axis, Substitute A for x max in equation (2), P E = 1 2 k A 2 (5) According to the conservation of energy, the initial kinetic energy is same as the elastic potential energy of the spring. Equate equations (4) and (5), K E = P E Substitute 1 2 m v max 2 for KE and 1 2 k A 2 for PE , 1 2 m v max 2 = 1 2 k A 2 m v max 2 = k A 2 A 2 = m v max 2 k A = m v max 2 k A = v max m k (6) After the block splits into two identical parts, its mass become half of the mass of the original block, that is, m 2 . Consider the amplitude of the split block to be A ′ , velocity to be v ′ , kinetic energy to be K E ′ and potential energy to be P E ′ . Value of the kinetic energy of the split block is, Substitute m 2 for m , K E ′ for KE , v ′ for v in equation (1), K E ′ = 1 2 ( m 2 ) ( v ′ ) 2 (7) Substitute P E ′ for PE and A ′ for A in equation (5), P E ′ = 1 2 k ( A ′ ) 2 (8) Equate equations (7) and (8). K E ′ = P E ′ Substitute 1 2 ( m 2 ) ( v ′ ) 2 for K E ′ and 1 2 k ( A ′ ) 2 for P E ′ , 1 2 ( m 2 ) ( v ′ ) 2 = 1 2 k ( A ′ ) 2 (9) Since, the block is split when it was having maximum speed, the speed of the half of the block is the maximum speed attained by the original block, that is, v ′ = v max (b) To determine Magnitude of the amplitude and frequency of the simple harmonic motion.

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ISBN: 9781118486894

Author: David Young, Shane Stadler

Publisher: WILEY

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Chapter 10, Problem 39P

(a)

To determine

The amplitude and frequency of the simple harmonic motion.

(b)

To determine

Magnitude of the amplitude and frequency of the simple harmonic motion.

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Chapter 10, Problem 38P

Chapter 10, Problem 40P

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