A particle executes simple harmonic motion and is located at x=a, b, and c at times t0, 2t0, and 3t0 respectively. The frequency of the oscillation is?

a=A cos ωt0(1)
b=A cos 2ωt0 (2)
c=A cos 3ωt0 (3)
On adding (1) and (3)
a+c=A(cos ωt0+cos 3ωt0)
a+c=2Acos(3ωt0+ωt0/2)cos(3ωt0−ωt0/2)
a+c=2Acos 2ωt0 cos ωt0
from (2), b=A cos 2ωt0
a+c=2b cos ωt0
cos⁻¹(a+c/2b)=ωt0
ω=cos⁻¹(a+c/2b)/t0
f=ω/2π=cos⁻¹(a+c/2b)/2πt0
Therefore, the frequency is proportional to cos⁻¹(a+c/2b)