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An object of mass 0.2 kg executes simple...

An object of mass 0.2 kg executes simple harmonic oscillation along the x-axis with a frequency `(25)/pi`. At the position x = 0.04m, the object has kinetic energy 0.5J and potential energy 0.4J. amplitude of oscillation is (potential energy is zero mean position).

A

0.05

B

0.06

C

0.01

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`E = (1)/(2) m omega^(2) ^^^(2)`
`rArr E = (1)/(2) m (2pi f)^(2) A^(2)`
`rArr A = (1)/(2pi f) sqrt((2E)/(m))`
Putting `E = K + U` we obtain `A = (1)/(2pi (25//pi)) sqrt((2 xx (0.5 + 0.4))/(0.2))`
`rArr A = 0.06m`
Hence B is correct
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