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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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Answer» For any SHM KE when particle is at X from mean position = 1/2 mw2(A2 -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 W w |
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