Question

Which of the following describe for the SHM of Fig.:

(a) $\mathbf{-}\mathit{\pi }\mathbf{<}\mathit{\varphi }\mathbf{<}\mathbf{-}\mathit{\pi }\mathbf{/}\mathbf{2}$,

(b) $\mathbf{}\mathit{\pi }\mathbf{<}\mathit{\varphi }\mathbf{<}\mathbf{3}\mathit{\pi }\mathbf{/}\mathbf{2}$,

(c) $\mathbf{-}\mathbf{3}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{<}\mathit{\varphi }\mathbf{<}\mathbf{-}\mathit{\pi }$?

Short answer

The limits which describes $\varphi$for the SHM are $-\pi <\varphi <-\frac{\pi }{2}\mathrm{and}\pi <\varphi <\frac{3\pi }{2}$.

Step-by-step solution

The given data 

The graph of x versus t is given.

Understanding the concept of SHM of a particle

We can use the displacement of SHM to find the expression for X at t = 0. Then, by observing the graph we can interpret the possible values of the phase angle.

Formula:

The displacement equation of a particle in motion,$X(\mathrm{t})=A\cos (\omega t+\varphi )$ (i)

where, A is the maximum displacement, $\omega$is the angular velocity, t is time and $\varphi$is the phase difference.

Calculation of the phase angle that describes the SHM

At t = 0 we get, we can get the displacement of the particle using equation (i) as follows:

$X=A\cos \left(\varphi \right)$

From the graph, we can see that at t = 0 the value of X is negative; we know that the value of cosine function is negative in II and III quadrant.

Hence, the limits of the phase are $-\pi <\varphi <-\frac{\pi }{2}\mathrm{and}\pi <\varphi <\frac{3\pi }{2}$.