Question

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

Answer 1

Correct Answer - 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))`