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Question

(A) The equation of motion for a particle in $SHM$ starting from the mean position is $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.At $x = 0

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$T_1 < T_2$

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(A) The equation of motion for a particle in $SHM$ starting from the mean position is $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.At $x = 0$,the time $t_0 = 0$.At $x = A/2$,we have $\frac{A}{2} = A \sin(\omega t_1)$,which gives $\sin(\omega t_1) = 1/2$. Thus,$\omega t_1 = \pi/6$,so $t_1 = \frac{T}{12}$.Therefore,$T_1 = t_1 - t_0 = \frac{T}{12}$.At $x = A$,we have $A = A \sin(\omega t_2)$,which gives $\sin(\omega t_2) = 1$. Thus,$\omega t_2 = \pi/2$,so $t_2 = \frac{T}{4}$.Therefore,$T_2 = t_2 - t_1 = \frac{T}{4} - \frac{T}{12} = \frac{3T - T}{12} = \frac{2T}{12} = \frac{T}{6}$.Comparing the two,$T_1 = \frac{T}{12}$ and $T_2 = \frac{T}{6}$.Since $\frac{T}{12} < \frac{T}{6}$,we conclude that $T_1 < T_2$.

$A$ particle executes simple harmonic motion between $x=-A$ and $x=+A$. If it takes a time $T_1$ to go from $x=0$ to $x=A/2$ and $T_2$ to go from $x=A/2$ to $x=A$,then:

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