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A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

```

5 years ago

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```										A = 10cm =0.1m
kinetic energy = mA2w2 cos2wt/2 ................1
potential energy = mA2w2sin2wt/2 ..................2
PE = KE           (GIVEN)
so
tanwt = 1
sinwt =1/sqrt(2)
now ,  x = Asinwt
=A/sqrt2= 7cm
```
5 years ago

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