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A particle is executing simple harmonic motion along a straight line with amplitude A. Its potential energy is changing with time according to the equation $U = 2 (1 - \cos 80 πt) J$ .Then frequency of SHM and maximum kinetic energy of the particle are respectively:

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20 Hz, 4 J

40 Hz, 2 J

80 Hz, 4 J

60 Hz, 4 J

answer is A.

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Detailed Solution

f = frequency of potential energy $= \frac{80 π}{2 π} Hz = 40 Hz$

$∴$ Frequency of SHM, $f' = \frac{f}{2} = 20 Hz$

$U_{\max} = 2 [1 - (- 1)] J = 4 J, U_{\min} = 2 [1 - 1] = 0$

$∴ K_{\max} = U_{\max} - U_{\min} = (4 - 0) J = 4 J$

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