Question

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?

Answer 1

The value of the distance is
`\bf{14.52~cm}`.

Step-by-step explanation:

The velocity of a particle(v) executing SHM is

`v = \omega \sqrt{A^(2) - x^(2)}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)`

where,
`\omega` is the angular frequency,
`A` is the amplitude of the oscillation and
`x` is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e.,
`x = 0`.

The maximum velocity(
`\bf{v_(m)}`) is

`v_(m) = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)`

Divide equation (1) by equation(2).

`(v)/(v_(m)) = \frac{\sqrt{A^(2) - x^(2)}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)`

Given,
`v = 0.25 v_(m)` and
`A = 15~cm`. Substitute these values in equation (3).

`&& (1)/(4) = \frac{\sqrt{15^(2) - x^(2)}}{15}\\&or,& A = 14.52~cm`