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A body is performing simple harmonic mot...

A body is performing simple harmonic motion. Then its

A

average total energy per cycle is equal to its maximum kinetic energy

B

average kinetic energy per cycle is equal to half of its maximum kinetic energy

C

mean velocity over a complete cycle is equal to `(2)/(pi)` times of its maximum velocity

D

root mean square velocity is `(1)/(sqrt(2))` time of its maximum velocity

Text Solution

Verified by Experts

The correct Answer is:
A, B, D

In SHM, total energy per cycle `=(1)/(2) m omega^(2)r^(2)`
Max. K.E. `=(1)/(2)mV_(max)^(2)=(1)/(2)m(romega)^(2)=(1)/(2)momega^(2)r^(2)`
Min. KE `(1)/(2) m(0)^(2)=0` lt brgt Average KE per cycle `=((1)/(2)momega^(2)r^(2)+0)/(2)=(1)/(4)momega^(2)r^(2)=(1)/(2)(KE)_(max)`
Max. velocity `=romega,` Min velocity `=0`
Mean velocity `=(romega+0)/(2)=(1)/(2)romega=(1)/(2)(romega)=(1)/(2)max. velocity`
rms velocity `=sqrt((0+r^(2)omega^(2))/(2))=(romega)/(sqrt(2))=(1)/(sqrt(2))(upsilon_(max))`
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