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Q.

A particle is moving on X-axis and has potential energy U=2−20x+5x2 joule, where x is position. The particle is released at x = -3. If the mass of the particle is 0.1 kg, then the maximum velocity (in ms-1) of the particle is 25β. If amplitude is 5 m, then the value of β is __________.

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answer is 2.

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Detailed Solution

U=2−20x+5x2dUdx=−20+10xF=−dUdx=20−10x=−10(x−2)∴The particle is executing SHM about mean position x - 2 = 0 or x = 2∴ k=10 ⇒ mω2=10 ω2=10m=100.1=100⇒ ω=10rads−1 vmax=Aω=5(10)=50ms−1∴ β=2

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