Q.

A particle of mass 10 g is placed in- la- potential field given by V=50x2+100 J/kg. If the frequency of oscillation is found to be xπcycle/s, the value of x is _______.

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answer is 5.

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Detailed Solution

Given that potential energy is U=mV⇒ U=50x2+10010−2F=−dUdx=−(100x)10−2⇒ mω2x=−100×10−2x10×10−3ω2x=100×10−2x⇒ ω2=100⇒ ω=10rad/s⇒ f=ω2π=10π=5πHence x = 5.
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