A particle executes simple harmonic motion between x = - A and x = + A. The time taken for it to go fromO to A /2 is T1 and to go from A/2 to A is T2, then

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 T1 and to go from A/2 to A is T2, then (A)  T1 < T2 (B)  T1 > T2 (C)  T1 = T2 (D)  T1 = 2 T2

Tardigrade Admin · Accepted Answer

. 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} < T 2

NOTE
From the equations of SHM we can show that

t_{1} = T_{0 - A /2} = T /12 an d t_{2} = T_{A /2 - A} = T /6

So, that

t_{1} + t_{2} = T_{0 - A} = T /4

Q. 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 $A /2 t o A i s T_{2},$ then

$T_{1} < T_{2}$

$T_{1} > T_{2}$

$T_{1} = T_{2}$

$T_{1} = 2 T_{2}$

Q. A particle executes simple harmonic motion between x = - A and x = + A. The time taken for it to go fromO to A /2 is T1​ and to go from A/2toAisT2​, then

T1​<T2​

T1​>T2​

T1​=T2​

T1​=2T2​

Q. 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 $A /2 t o A i s T_{2},$ then

$T_{1} < T_{2}$

$T_{1} > T_{2}$

$T_{1} = T_{2}$

$T_{1} = 2 T_{2}$