A particle executes SHM with amplitude $a$ and time period $T$ . The displacement of the particle from the mean position when its speed is half of maximum speed is $\frac{\sqrt{x} a}{2}$ . The value of $x$ is _______.

Open in App

Solution

For a particle executing SHM, $v = ω \sqrt{A^{2} - X^{2}}$

$\Rightarrow \frac{ω a}{2} = ω \sqrt{a^{2} - X^{2}}$

$\Rightarrow \frac{ω^{2} a^{2}}{4} = ω^{2} a^{2} - ω^{2} X^{2}$

$\Rightarrow X^{2} = \frac{3}{4} a^{2}$

$\Rightarrow X = \frac{\sqrt{3} a}{2} = \frac{\sqrt{x} a}{2}$

$\Rightarrow x = 3$

Suggest Corrections

Similar questions

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. For a particle executing SHM having amplitude 'a' the speed of the particle is one half of its maximum speed when its displacement from the mean position is

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

Q. A particle executes a SHM of time period T. The time taken by the particle to go directly from its mean position to half the amplitude.

Join BYJU'S Learning Program

A particle executes SHM with amplitude a and time period T. The displacement of the particle from the mean position when its speed is half of maximum speed is √xa2. The value of x is _______.

For a particle executing SHM, v=ω√A2−X2 ⇒ωa2=ω√a2−X2 ⇒ω2a24=ω2a2−ω2X2 ⇒X2=34a2 ⇒X=√3a2=√xa2 ⇒x=3

A particle executes SHM with amplitude $a$ and time period $T$ . The displacement of the particle from the mean position when its speed is half of maximum speed is $\frac{\sqrt{x} a}{2}$ . The value of $x$ is _______.