Question

A particle executes simple harmonic motion between x = - A and x = + A .The time taken by it to go from O to A /2 is T${}_1$ and to go from A /2 to A is T${}_2$. Then

• A. T${}_1$ > T${}_2$
• B. T${}_1$< T${}_2$
• C. T${}_1$ = T${}_2$
• D. T${}_1$ = 2 T${}_2$

Answer 1

Answer: (B)

The correct option is A T${}_1$< T${}_2$
(a) The equation of SHM, when it starts swinging from the mean position, is $y=Asin\omega t$ .

At mean position, y=0, t=0 .....(1)

$\frac{A}{2}=Asin\omega t_2$ or $sin\omega t_2=\frac{1}{2}=sin\frac{\pi }{6}$

$\omega t_2=\frac{\pi }{6}$ or $t_2=\frac{\pi }{6\omega }$ .....(2)

At y=A time for travel from 0 to A=$t_3$

A=Asin$\omega t_3$

or sin$\omega t_3=1=sin\frac{\pi }{2}$ or $t_3=\frac{\pi }{2\omega }$ .....(3)

$T_1=t_2-t_1=\frac{\pi }{6\omega }$ .....(4)

$T_2=t_3-t_2=\frac{\pi }{2\omega }-\frac{\pi }{6\omega }=\frac{\pi }{3\omega }$ .....(5)

$\frac{T_1}{T_2}=\frac{\frac{\pi }{6\omega }}{\frac{\pi }{3\omega }}=\frac{3\omega }{6\omega }=\frac{1}{2}$ or $\frac{T_1}{T_2}=\frac{1}{2}$

$T_1<T_2$