A particle is executing SHM with a period π sec. When it is passing through the center of its path, its velocity is 0.1 m/s. What is its velocity when it is at a distance of 0.03 m from the mean position?

Question

A particle is executing SHM with a period π sec. When it is passing through the center of its path, its velocity is 0.1 m/s. What is its velocity when it is at a distance of 0.03 m from the mean position? (A) 0.08 m/s (B) 0.05 m/s (C) 0.04 m/s (D) 0.01 m/s

Tardigrade Admin · Accepted Answer

When the point is at a distance from the mean position its, velocity is given by

v = ω a^{2} - y^{2}

Given,

T = \frac{2 π}{ω} = π

∴ ω = 2 5^{- 1}

and

v = aω 0.1 = a \times 2

a = \frac{0.1}{2} = 0.05 m y = 0.05 m

∴

Velocity,

v = 2 (0.05)^{2} - (0.03)^{2}

= 0.08 m / s

Q. A particle is executing SHM with a period πsec. When it is passing through the center of its path, its velocity is 0.1m/s. What is its velocity when it is at a distance of 0.03m from the mean position?

0.08 m/s

0.05 m/s

0.04 m/s

0.01 m/s

When the point is at a distance from the mean position its, velocity is given by v=ωa2−y2​ Given, T=ω2π​=π ∴ω=25−1 and v=aω0.1=a×2 a=20.1​=0.05my=0.05m ∴ Velocity, v=2(0.05)2−(0.03)2​ =0.08m/s

Q. A particle is executing SHM with a period $π sec$ . When it is passing through the center of its path, its velocity is $0.1 m / s$ . What is its velocity when it is at a distance of $0.03 m$ from the mean position?

When the point is at a distance from the mean position its, velocity is given by $v = ω a^{2} - y^{2}$
Given, $T = \frac{2 π}{ω} = π$
$∴ ω = 2 5^{- 1}$ and $v = aω 0.1 = a \times 2$
$a = \frac{0.1}{2} = 0.05 m y = 0.05 m$
$∴$ Velocity, $v = 2 (0.05)^{2} - (0.03)^{2}$
$= 0.08 m / s$