Question

A particle executes SHM with amplitude $0.4m$ and time period $24s$. The distance travelled by the particle in $10s$ is (initially particle is at its mean position)​

Answer 1

Step 1: Given data

Amplitude, $A=0.4m$

Time period, $T=24s$

$t=10s$

Step 2: To find

The distance traveled by the particle in $t=10s$.

Step 3: Calculating distance traveled

The equation for simple harmonic motion,

$x=A\sin \left(\omega t+\varphi \right)-------\left(1\right)$

Where $A$ is amplitude, $\omega$ is angular velocity, $\varphi$ is phase shift and $t$ is time.

We know, angular velocity

$$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{T} \\ \Rightarrow \omega =\frac{2\mathrm{\pi }}{24} \\ \Rightarrow \omega =\frac{\mathrm{\pi }}{12} \end{gathered}$$

$$\begin{gathered} Fromequation\left(1\right)\Rightarrow \\ x=A\sin \left(\frac{\mathrm{\pi }}{12}t\right)\left(\because \omega =\frac{\mathrm{\pi }}{12}\right) \\ x=0.4\sin \left(\frac{\mathrm{\pi }}{12}\times 10\right)\left(\because A=0.4m,t=10s\right) \\ x=0.4\sin \left(\frac{5\mathrm{\pi }}{6}\right) \\ x=0.4\times \frac{1}{2} \\ x=0.2m \end{gathered}$$

Therefore, The distance traveled by the particle in $t=10s$ is $0.2m$.