Question 18: Oscillations

Question

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec . The time required by it to move from $x=A$ to $x=\frac{A}{\sqrt{2}}$ is $\_\_\_\_$ sec.

Accepted Answer

For a particle starting from the extreme position $(x=A)$ at $t=0$, the displacement $x$ at any time $t$ is given by:
$$ x=A \cos (\omega t) $$
Where $\omega$ is the angular frequency, defined as $\omega=\frac{2 \pi}{\mathrm{~T}}$.
The given time period is $\mathrm{T}=5 \mathrm{sec}$ :
$$ \omega=\frac{2 \pi}{5} \mathrm{rad} / \mathrm{s} $$
The particle moves to $\mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}$. Substituting into the displacement equation:
$$ \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \cos \left(\frac{2 \pi}{5} \mathrm{t}\right) $$
$$\Rightarrow $$ $$ \frac{1}{\sqrt{2}}=\cos \left(\frac{2 \pi}{5} t\right) $$
Since $\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ :
$$ \frac{\pi}{4}=\frac{2 \pi}{5} \mathrm{t} $$
$$\Rightarrow $$ $$ \mathrm{t}=\frac{5}{8} \mathrm{sec} $$
Therefore, the correct option is (C).

Answer

1/4

Answer

5/4

Answer

5/8

Answer

3/8

Oscillations

A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec . The time required by it to move from x=A to x=\frac{A}{\sqrt{2}} is \_\_\_\_ sec.

Choose the correct answer: