Question

A particle executes simple harmonic motion between x = - A and x = + A. The time taken for it to go fromO 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: (A)

$T_1 < T_2$

Answer 2

. In SHM, velocity of particle also oscillates simple
harmonically. Speed is more near the mean position and
less near the extreme positions. Therefore, the time taken

for the particle to go from O to A/2 will be less than the

time taken to go it from A/2 to A, or$T_1 < T2$

NOTE

From the equations of SHM we can show that

$t_1 =T_{0-A/2} =T/ 12 \, \, and \, \, t_2 = T_ { A/2 -A } = T/6$
So, that $t_1 +t_2 = T_{0-A } = T/4$