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

Answer

A

Explanation

Let x coordinate of the particle be given by `x = A sin omega`. With given information we can write the following :
`a = A sin omegat_(0), b = A sin 2omegat_(0), c = A sin 3omegat_(0)`
`a + c = A[sin omegat_(0) + sin 3omegat_(0)] = 2 A sin 2omegat_(0) cos omegat_(0)`
`(a + c)/(b) = 2 cos omegat_(0)`
` omega = (1)/(t_(0)) cos^(-1)((a + c)/(2b)) rArr f = (1)/(2pit_(0)) cos^(-1) ((a + c)/(2b))`