Question

A particle executes simple harmonic motion and is located at x = a, b, and c at times t₀, 2t_0, and 3t₀ respectively. The frequency of the oscillation is ______.

Options

• `1/2pi"t"_0 cos^-1((a + c)/(2b))`

• `1/2pi"t"_0 cos^-1((a + c)/(2c))`

• `1/2pi"t"_0 cos^-1((2a + 3c)/(b))`

• `1/2pi"t"_0 cos^-1((a + 2b)/(2b))`

Answer 1

A particle executes simple harmonic motion and is located at x = a, b, and c at times t₀, 2t_0, and 3t₀ respectively. The frequency of the oscillation is `1/2pi"t"_0 cos^-1((a+c)/(2b))`. 

Explanation:

We have

a= A sin `omega`t₀; b= A sin 2`omega`t₀ and c= A sin 3`omega`t₀

Now,

a + c = A sin `omega`t0 + A sin 3`omega`t₀

∴ a + c = 2 A sin 2`omega`t0 cos `omega`t0

⇒ a + c =2 b cos `omega`t0

∴ `omega"t"_0 = cos^-1 ((a + c)/(2b))`

⇒ v = `1/2pi"t"_0 cos ^-1 ((a + c)/(2b))`