A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is:

Question

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is: (A) √3 cm  (B) √5 cm (C) 2 √3 cm (D) 2 √5 cm

Tardigrade Admin · Accepted Answer

A ω=10 A=4 so ω= (5/2) v=ω √A2-x2=2 cm / s

Q. A particle executes simple harmonic motion with an amplitude of $4 c m$ . At the mean position the velocity of the particle is $10 c m / s$ . The distance of the particle from the mean position when its speed becomes $5 c m / s$ is:

$3 c m$

$5 c m$

$23 c m$

$25 c m$

$A ω = 10& A = 4$
so $ω = \frac{5}{2}$
$v = ω A^{2} - x^{2} = 2 c m / s$

Q. A particle executes simple harmonic motion with an amplitude of 4cm. At the mean position the velocity of the particle is 10cm/s. The distance of the particle from the mean position when its speed becomes 5cm/s is:

3​cm

5​cm

23​cm

25​cm