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For the spring pendulum shown in Fig., the value of spring constant is $3 \times 10^{4} N/m$ and amplitude of oscillation is 0.1m. The total mechanical energy of oscillating system is 200J. Mark out the correct option (s)

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Minimum PE of the oscillating system is 50J

Maximum PE of the oscillating system is 200J

Maximum KE of the oscillating system is 200J

Minimum KE of the oscillating system is 150J

answer is A, B, D.

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Detailed Solution

K = 3 \times 10^{4} N/m; A = 0.1 m; ME = 200 J

${KE}_{\max} = \frac{1}{2} {kA}^{2} = 150$

$∴ {PE}_{min} = ME - {KE}_{max}$

At extreme position KE = 0

$∴ {PE}_{min} = PE = ME$

$∴ ME = {PE}_{\max} = 200 J$

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