Question

A particle executes simple harmonic motion and is located at x=a, b and c at time to,2to, and 3to respectively.
The frequency of the oscillation is

1.) 1/2πto cos-¹ [a+c/2b]
2.) 1/2πto cos-¹[a+b/2c]
3.) 1/2πto cos-¹ [2a+3c/b]
4.) 1/4πto cos-¹[a+2b/3c]

Answer 1

● Answer-
(1) 1/2πt0 cosinv[(a+c)/2b]

● Explaination-
Let,
a = Asin(wt0)
b = Acos(2wt0)
c = Acos(3wt0)

Now take,
a + c = Asin(wt0) + Asin(3wt0)
a + c = 2Asin(2wt0).cos(wt0)
a + c = 2b.cos(wt0)
cos(wt0) = (a+c)/2b

Taking inverse on both sides,
wt0 = cosinv[(a+c)/2b]
2πft0 = cosinv[(a+c)/2b]
f = cosinv[(a+c)/2b] / 2πt0

Therefore, frequency of oscillation is 1/2πt0 cosinv[(a+c)/2b] .

Hope this helps you...