Exam

BITSAT

Year

2024

Paper

Paper 2

Subject

Physics

Chapter

Oscillations

BITSAT 2024 · Paper 2 · Physics · Oscillations

A particle executes simple harmonic motion between x=-A and x=+A ....

A particle executes simple harmonic motion between $x=-A$ and $x=+A$. If time taken by particle to go from $x=0$ to $\frac{A}{2}$ is $2 s$; then time taken by particle in going from $x=\frac{A}{2}$ to $A$ is:

Step-by-step solution

Let time from 0 to $\mathrm{A} / 2$ is $\mathrm{t}_1$ and from $\mathrm{A} / 2$ to A is $\mathrm{t}_2$ From the standard equation of SHM, $\Rightarrow \frac{\mathrm{A}}{2}=\mathrm{A} \sin \left(\omega \mathrm{t}_1\right)\Rightarrow \omega t_1=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$....(i) Using $x=A_0 \sin \omega t$ again $A=A \sin \omega\left(t_1+t_2\right)\omega\left(t_1+t_2\right)=\sin ^{-1}(1)=\frac{\pi}{2}$ Using (i) $\omega t_2=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$.....(ii) Dividing equation (i) by (ii) we get $\frac{t_1}{t_2}=\frac{1}{2}\Rightarrow \quad t_2=2 t_1=2 \times 2=4 \mathrm\sec$