Answer to Question #89818 in Mechanics | Relativity for Sumanth

Question

A particle execute SHM between x =-A and x=+A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2, then find relation between t1 and t2

Answer 1

Without loss of generality, let us assume that at time t = 0 an oscillator passes the equilibrium position (x = 0). Hence, its law of motion has the following form:

$$x(t) = A \sin{\omega t}$$

We observe the considered positions (A/2 and A) of the oscillator at the moments of time:

$$x(t_1) = \frac{A}{2} = A \sin{\omega t_1} \, \Rightarrow \, \sin{\omega t_1} = \frac{1}{2} \, \Rightarrow \, \omega t_1 = \frac{\pi}{6}$$

and

$$x (t_A) =A = A \sin{\omega t_A} \, \Rightarrow \, \sin{\omega t_A} = 1 \, \Rightarrow \, \omega t_A = \frac{\pi}{2}$$

The time t₂ is just the difference between t_A and t₁:

$$t_2 = t_A - t_1 \, \Rightarrow \, \omega t_2 = \omega t_A - \omega t_1 = \frac{\pi}{3}$$

Finally,

$$\frac{t_2}{t_1} = \frac{\omega t_2}{\omega t_1} = \frac{\pi/3}{\pi/6} = 2$$

Answer: the ratio between t₂ and t₁ is 2.