Question

a particle of mass 100gm performs linear SHM along a path of length 20cm with a frequency of 60hz .Find the value of momentum when it is at a distance of 2cm from positive extrimity?

Answer 1

Answer on Question #58378, Physics / Mechanics | Relativity

A particle of mass 100g performs linear SHM along a path of length 20cm with a frequency of 60Hz. Find the value of momentum when it is at a distance of 2cm from positive extremity?

Solution:

Let:

- v is the velocity at displacement x from the mid-point,

- $f = 60\,\mathrm{Hz}$ is the frequency,

- k is the spring constant,

- $m = 100\,\mathrm{g}$ is the mass,

- $A = 10\,\mathrm{cm}$ is the amplitude.

An object experiencing simple harmonic motion is traveling in one dimension, and its one-dimensional motion is given by an equation of the form

$$x = A \cos \omega t$$

The velocity is given by

$$v = \omega A \cos \omega t$$

From first equation

$$\cos \omega t = \frac{x}{A} = \frac{2\,\mathrm{cm}}{10\,\mathrm{cm}} = 0.2$$

Thus,

$$v = \omega A \cos \omega t = 2\pi f A \cos \omega t = 2 \cdot \pi \cdot 60 \cdot 0.1 \cdot 0.2 = 7.54\,\mathrm{m/s}$$

The momentum is

$$p = m v = (0.1\,\mathrm{kg}) \cdot (7.54\,\mathrm{m/s}) = 0.754\,\mathrm{kg} \cdot \mathrm{m/s}$$

Answer: $0.754\,\mathrm{kg} \cdot \mathrm{m/s}$

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