91Ó°ÊÓ

The displacement in simple harmonic motion as a function of time is given by

$\mathbf{x}\mathbf{=}\mathbf{Acos}\mathbf{(}\mathbf{Ó\neg ³Ù}\mathbf{-}\mathbf{Ï•}\mathbf{)}$ …â¶Ä¦â¶Ä¦â¶Ä¦.(1)

Here given that the period of SHM is T =0.3 s.

Amplitude of SHM is $\mathrm{A}=6\mathrm{cm}=6×{10}^{-2}\mathrm{m}$.

As the displacement in simple harmonic motion as a function of time is given by

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

Since at time t =0 s, the object is instantaneously at rest at x=6 cm.

Hence at t =0 s, the phase angle $\mathrm{Ï•}=0$.

So, from equation (1),

$\mathrm{x}=\mathrm{Acos}\left(\mathrm{Ó\neg ³Ù}\right)$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(2)

Since the angular frequency $\mathrm{Ó\neg }=2\mathrm{Ï„Ï„´Ú}=\frac{2\mathrm{Ï„Ï„}}{\mathrm{T}}$, hence

$\mathrm{x}=\mathrm{A}\cos \left(\frac{2\mathrm{Ï„Ï„}}{\mathrm{T}}\mathrm{t}\right)$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(3)

Therefore, from equation (3), the time taken by the object to go from x=6 cm to x=-1.5 cm will be

$$\begin{gathered} t=\left(\frac{0.3s}{2\mathrm{Ï„}\mathrm{Ï„}}\right){\cos }^{-1}\left(-0.25\right) \\ =0.087s \end{gathered}$$

Hence, time takes the object to go from$\mathrm{x}=6.00\mathrm{cm}\mathrm{to}\mathrm{x}=-1.50\mathrm{cm}$ at t =0.087 s.