A particle is executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 and its acceleration is 0.5 m / s 2. The maximum velocity of the particle is ( in m / s )

Question

A particle is executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 and its acceleration is 0.5 m / s 2. The maximum velocity of the particle is ( in m / s ) (A) 0.01 (B) 0.05 (C) 0.5 (D) 0.25

Tardigrade Admin · Accepted Answer

Maximum acceleration α=ω2 y ∴ 0.5=ω2 × 0.02 or ω2=(0.5/0.02)=25 So, ω=5 Now, maximum velocity is v=a ω=0.1 × 5=0.5 m / s

Q. A particle is executing simple harmonic motion with amplitude of $0.1 m$ . At a certain instant when its displacement is $0.02$ and its acceleration is $0.5 m / s^{2}$ . The maximum velocity of the particle is $(inm / s)$

0.01

0.05

0.5

0.25

Maximum acceleration
$α = ω^{2} y$
$∴ 0.5 = ω^{2} \times 0.02$
or $ω^{2} = \frac{0.5}{0.02} = 25$
So, $ω = 5$
Now, maximum velocity is
$v = aω = 0.1 \times 5 = 0.5 m / s$

Q. A particle is executing simple harmonic motion with amplitude of 0.1m. At a certain instant when its displacement is 0.02 and its acceleration is 0.5m/s2. The maximum velocity of the particle is (inm/s)

0.01

0.05

0.5

0.25

Maximum acceleration α=ω2y ∴0.5=ω2×0.02 or ω2=0.020.5​=25 So, ω=5 Now, maximum velocity is v=aω=0.1×5=0.5m/s

Q. A particle is executing simple harmonic motion with amplitude of $0.1 m$ . At a certain instant when its displacement is $0.02$ and its acceleration is $0.5 m / s^{2}$ . The maximum velocity of the particle is $(inm / s)$

Maximum acceleration
$α = ω^{2} y$
$∴ 0.5 = ω^{2} \times 0.02$
or $ω^{2} = \frac{0.5}{0.02} = 25$
So, $ω = 5$
Now, maximum velocity is
$v = aω = 0.1 \times 5 = 0.5 m / s$