Question

A particle is executing shm of amplitude r. At a distance s from the mean position, the particle receives a blow in the direction opposite to motion which instantaneously doubles the velocity find the new amplitude.

Answer 1

Using the formula of the velocity of the Amplitude,

Velocity , v² = w²(A² - s²), where A is the amplitude, and s is the distance of the particle from mean position.

In the Question, Amplitude(A) = r,

therefore,

v² = w²(r² - x²)

Let the velocity of the Particle when it is at a distance of s from mean position be u.

∴ u² = w²(A² - s²), Where A is the amplitude of the new S.H.M. and s is the distance of particle from mean position.

It is given, that velocity become double, which means, u = 2v

∴ u² = 4v²

w²(A² - s²) = 4w²(r² - s²)

A² - s² = 4(r² - s²)

A² = 4r² - 4s² + s²

A² = 4r² - 3s²

A = √(4r² - 3s²)

Hence, the new Amplitude of the S.H.M. is given by this formula, √(4r² - 3s²).

Hope it helps.

Answer 2

Explanation:

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