Question

An object in SHM oscillates with a period of $4.0s$and an amplitude of $10cm$. How long does the object take to move from $x=0.0cm$ to $x=6.0cm$?

Short answer

To move from $x=0.0cm$to $x=6.0cm$, the total time taken is $0.4s$.

Step-by-step solution

Given Information 

Time period, $T=4s$.

The amplitude,$A=10cm$.

:Formula used 

Equation of the simple harmonic motion is $x=A\sin \omega t$.

Where, at time $t$, $x$is the position, $A$is the amplitude and$\omega$is the angular frequency.

Calculation 

Substitute the values to obtain$t$.

$$\begin{gathered} 6=10\sin \left[\frac{2\mathrm{\pi }}{T}\times t\right] \\ \frac{6}{10}=\sin \left[\frac{2\mathrm{\pi }}{4}\times t\right] \\ {\sin }^{-1}\left(0.6\right)=\frac{\mathrm{\pi }}{2}\times t \\ t=0.4s \end{gathered}$$

Final Answer 

To move from $x=0.0cm$to $x=6.0cm$, the total time taken is $0.4s$.