A particle is describing S.H.M with displacement amplitude A. At time t = 0 its speed is half of the maximum speed. At time t = 0 its displacement w.r.t the mean position is

\frac{3}{2} A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{\sqrt{3}}{2} A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{\frac{3}{2}} A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{\sqrt{3}}{2} A$
$v = ω \sqrt{A^{2} - X^{2}}, a n d v = \frac{v_{m a x}}{2} = \frac{A ω}{2}$ .Substituting in the equation we get $x = \frac{\sqrt{3}}{2} A$

Suggest Corrections

Similar questions

Q. A particle execute S.H.M from the mean position. its amplitude is

A

, its time period is "T'. At what displacement,its speed is half of its maximum speed.

A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is

[Haryana CEE 1996; CBSE PMT 1996; MH CET 2002]

Q. A particle execute

S H M

from the mean position

.

its amplitude is

A

its time period

T .

At what displacement its speed is half of its maximum speed

?

Q. At t = 0, the displacement of a particle in S.H.M. is half its amplitude. Its initial phase is :

Join BYJU'S Learning Program

A particle is describing S.H.M with displacement amplitude A. At time t = 0 its speed is half of the maximum speed. At time t = 0 its displacement w.r.t the mean position is

The correct option is C √32Av=ω√A2−X2,andv=vmax2=Aω2 .Substituting in the equation we get x=√32A

The correct option is C $\frac{\sqrt{3}}{2} A$
$v = ω \sqrt{A^{2} - X^{2}}, a n d v = \frac{v_{m a x}}{2} = \frac{A ω}{2}$ .Substituting in the equation we get $x = \frac{\sqrt{3}}{2} A$