A particle is executing simple harmonic motion (S.H.M.). Its acceleration at a distance of 1 cm from the mean position is 3 cm s^(-2). If its velocity…

Question

A particle is executing simple harmonic motion (S.H.M.). Its acceleration at a distance of 1 cm from the mean position is 3 cm s^(-2). If its velocity is 6 cm s^(-1) when it is at a distance of 2 cm from its mean position, then the amplitude of S.H.M. is,

MARKS Content Team · Accepted Answer

Correct answer: (B) 4 cm. Recall the proper formula of acceleration and velocity in terms of amplitude and angular velocity, a=-ω^2x and v=ω√(A^2-x^2), given, x=1 cm & a=3 cm s^(-2), it means the value of, ω=√ax=√31 , and the velocity, v=ω√(A^2-x^2)⇒6=√3√(A^2-2^2)⇒36=3A^2-4⇒A=4 cm

A particle is executing simple harmonic motion (S.H.M.). Its acceleration at a distance of $1 cm$ from the mean position is $3 cm s^{- 2} .$ If its velocity is $6 cm s^{- 1}$ when it is at a distance of $2 cm$ from its mean position, then the amplitude of S.H.M. is,

Solution

Relevant Questions

A particle is executing simple harmonic motion (S.H.M.). Its acceleration at a distance of 1 cm from the mean position is 3 cm s-2. If its velocity is 6 cm s-1 when it is at a distance of 2 cm from its mean position, then the amplitude of S.H.M. is,

Solution

Relevant Questions

A particle is executing simple harmonic motion (S.H.M.). Its acceleration at a distance of $1 cm$ from the mean position is $3 cm s^{- 2} .$ If its velocity is $6 cm s^{- 1}$ when it is at a distance of $2 cm$ from its mean position, then the amplitude of S.H.M. is,