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A certain simple harmonic vibrator of mass 0.1 kg has a total energy of 101. its displacement from the man position is 1 cm when it has equal kinetic and potential energies. the amplitude and frequency of vibration of the vibrator are.

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by (638 points)
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Best answer

For any SHM 

KE when particle is at X from mean position = 1/2 mw2(A-x2) and 

Potential energy = 1/2 mw2x2

KE=PE when  X= A/√2 =0.707 A 

A= amplitude

At X=1 cm KE and PE are equal mean

A= X.√2= 1.414 cm

Total energy E=1/2mw2A2 

W2= 2E/(mA2)=2×101/0.1x0.01m)

W2= 202000

W= 449.3 /s

w

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by (638 points)
For simplicity I have consider A= 0.01m instead of 0.014 m

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