The maximum potential energy of a block executing simple harmonic motion is 25 J. is the amplitude of oscillation. At , the kinetic energy of the block is [2023]

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Solution

Q.
The maximum potential energy of a block executing simple harmonic motion is 25 J. $A$ is the amplitude of oscillation. At $\frac{A}{2}$ , the kinetic energy of the block is [2023]

1 37.5 J

2 9.75 J

3 18.75 J

4 12.5 J

Ans.
(3)

$u_{\max} = \frac{1}{2} m ω^{2} A^{2} = 25 J$

$KE at \frac{A}{2} = \frac{1}{2} m v_{1}^{2} = \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{4})$

$KE = \frac{1}{2} m ω^{2} \frac{3 A^{2}}{4} = \frac{3}{4} (\frac{1}{2} m ω^{2} A^{2})$

$KE = \frac{3}{4} \times 25 = 18.75 J$

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