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Q.

For the spring pendulum shown in the given figure, the value of spring constant is 3 x 104 N/m and amplitude of oscillation is 0.1 m. The total mechanical energy of oscillating system is 200 J. Mark out the correct option(s).

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a

Minimum PE of the oscillating system is 50 J

b

Maximum PE of the oscillating system is 200 J.

c

Maximum KE of the oscillating system is 200 J.

d

Minimum KE of the oscillating system is 150 J.

answer is A.

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Detailed Solution

Total mechanical energy of the oscillating system is, E=Kmax+Umin  = Kmin+Umax  ,Kmin=0 at extreme position.So, Umax=E=200J       Kmax=mvmax22=m×A2ω22=KA22=150JSo, Umin=50J
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