91Ó°ÊÓ

Frequency (f) is the number of waves that pass through a point in a given period of time (T). It is inversely proportional to the time period.

$f=\frac{1}{T}$

The angular frequency, $\mathrm{Ó\neg }$is 2$\mathrm{Ï€}$times the frequency;

$\mathrm{Ó\neg }=2\mathrm{Ï€}f=\frac{2\mathrm{Ï€}}{T}\left(1\right)$

This equation implies that the time period in SHM does not depend on the amplitude of the SHM. So, if the amplitude is doubled, from x = 0.09 m to x = 0.18 m, the time taken by the block to travel from x = +0.18 m to x = -0.18 m will be the same t = 2.7 s as in the previous case in which the block is travelling from x = +0.09 m to x = -0.09m.

The displacement x as a function of time can be written as,

$x=A\cos \left(\mathrm{Ó\neg }t-\mathrm{Ï•}\right)\mathbf{}\mathbf{}\mathbf{}\left(2\right)$

Here, A is amplitude, $\mathrm{Ó\neg }$is angular frequency, t is time, and $\mathrm{Ï•}$is the phase angle.

If there is no phase, then equation (2) become,

$x=A\cos \left(\mathrm{Ó\neg }t\right)$

Here, the time period, $T_1$of the SHM is double of the time taken by the block to travel from x = +0.18 m to x = -0.18 m because this displacement is half of the complete cycle. Hence, localid="1668081609542" ${\mathit{T}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{}\mathit{s}$.

Now, equation (1) becomes,

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T} \\ =\frac{2\mathrm{Ï€}}{5.4s} \\ =1.16rad{s}^{-1} \end{gathered}$$

Say, at time the block is at x = +0.09 m, so from equation (2), you get,

$$\begin{gathered} 0.09m=\left(0.18m\right)×\cos \left[\left(1.16rad{s}^{-1}\right)t_1\right] \\ {t}_1=\frac{1}{1.16rad{s}^{-1}}×{\cos }^{-1}\left(0.5rad\right) \\ =\frac{1.047rad}{1.16rad{s}^{-1}} \\ =0.903s\left(3\right) \end{gathered}$$

Next consider that at time $t_2$the block is at x = -0.09 m, so from equation (2), you get,

$$\begin{gathered} -0.09m=\left(0.18m\right)×\cos \left(1.16rad{s}^{-1}\right)t_2] \\ {t}_1=\frac{1}{1.16ras{s}^{-1}}×{\cos }^{-1}\left(-0.5rad\right) \\ =\frac{2.094rad}{1.16rad{s}^{-1}} \\ =1.806s\left(4\right) \end{gathered}$$

So, from equations (3) & (4), you get the value of t, which is the time taken by the block to travel from x = +0.09 m to x = -0.09 m,

$$\begin{gathered} T_2=t_2-t_1 \\ =1.806s-0.903s \\ =0.903s \end{gathered}$$

Hence, the time taken by the block to travel from x = +0.09 m to x = -0.09 m is ${\mathit{T}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{903}\mathbf{}\mathit{s}$.