A pendulum is executing simple harmonic motion and its maximum kinetic energy is `K_(1)`. If the length of the pendulum is doubled and it perfoms simple harmonuc motion with the same amplitude as in the first case, its maximum kinetic energy is `K_(2)` Then:

Options

A. `K_(2)`
B. `K_(2)=2K_(1)`
C. `K_(2)=K_(1)`
D. `K_(2)=(k_(1))/4`

Answer

B

Explanation

Maximum kinetic energy `=1/2m omega^(2)A^(2)`
`omega=sqrt(g/L)`
`A=L theta`
`KE=1/2 m gL xxL^(2) theta^(2),KE=1/2mgltheta^(2)`
`K_(1)=1/2mgLtheta^(2)`
If length is double
`K_(2)=1/2mg (2L)theta^(2)` [Here we are assuming angular amplitude is same]
`(K_(1))/(K_(2))=(1/2mg tl theta^(2))/(1/2 mg (2L)theta^(2))=1/2`
`K_(2)=2K_(1)`