A particle executes simple harmonic motion and is located at ` x = a`, b at times `t_(0),2t_(0) and3t_(0)` respectively. The frequency of the oscillation is :

Options

A. `(1)/(2pi t_(0))cos^(-1)((a+b)/(2c))`
B. `(1)/(2pi t_(0))cos^(-1)((a+b)/(3c))`
C. `(1)/(2pi t_(0))cos^(-1)((2a+3c)/(b))`
D. `(1)/(2pi t_(0))cos^(-1)((a+c)/(2b))`

Answer

D

Explanation

Using y = A `sin omega t`
`a=A sin omega t_(0)`
`b=A sin 2omega t_(0)`
`c=A sin 3 omega t_(0)`
`a+c=A[sin omega t_(0)+A sin 3omega t_(0)]=2A sin 2 omega t_(0)cos omega t_(0)`
`(a+c)/(b)=2cos omega t_(0)`
`implies" "omega=(1)/(t_(0))cos^(-1)((a+c)/(2b))impliesf=(1)/(2pi t_(0))cos^(-1)((a+c)/(2b))`