An object is undergoing SHM with period 0.900 s and amplitude 0.320 m. At t = 0 the object is at x = 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from x = 0.320 m to x = 0.160 m and (b) from x = 0.160 m to x = 0.

The general equation for displacement in SHM as a function of time is,

$\mathit{x}\mathbf{=}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\left(\omega t-\varphi \right)....................\left(1\right)$

Where A is the amplitude, is angular frequency and is phase angle.

We know that,

$\mathit{\omega }\mathbf{=}\frac{\mathbf{2}\mathbf{\prod }_{}}{\mathbf{T}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}.................\left(2\right)$

Hence,

$$\begin{gathered} \omega =\frac{2\prod _{}}{0.9s} \\ \omega =6.98rad/s \end{gathered}$$

It was given that A=0.320 m. Consider phase angle $\varphi =0°$.

Therefore, final equation will be

$x=0.320m\times \cos \left[\left(6.98rad/s\right)t\right]...................\left(3\right)$

a)

From equation (3)

$$\begin{gathered} 0.16m=0.32m\times \cos \left[\left(6.98rad/s\right)t_1\right] \\ {t}_1=\frac{{\cos }^{-1}\left(0.5rad\right)}{6.98rad/s} \\ {t}_1=\frac{1.47rad}{6.98rad/s} \\ {t}_1=0.15s \end{gathered}$$

The time taken will be

$$\begin{gathered} T_1=t_1-t \\ {T}_1=0.15s-0s \\ {T}_1=0.15s \end{gathered}$$

b)

Again, from equation (3)

$$\begin{gathered} 0=0.16m\times \cos \left[\left(6.98rad/s\right)t_2\right] \\ {t}_2=\frac{{\cos }^{-1}\left(0rad\right)}{6.98rad/s} \\ {t}_2=\frac{1.571rad}{6.98rad/s} \\ {t}_2=0.225s \end{gathered}$$

Time taken to go from x=0.160 m to x=0,

$$\begin{gathered} T_2=t_2-t_1 \\ {T}_2=0.225s-0.15s \\ {T}_2=0.075s \end{gathered}$$

Hence, a) The time taken to go from x=0.320 m to x= 0.160 m is $T_1$= 0.15 s.

b) The time taken to go from x=0.160 m to x= 0 is $T_2=0.075s$.