Question

A particle is executing simple harmonic motion with an amplitude of 2 cm. At the mean position velocity of the particle is 10 m/s. The distance of the particle from the mean position when its speed becomes $5\sqrt{2}$ m/s is

Answer 1

The particle is $\sqrt{2}\, \text{cm}$ (approximately 1.414 cm) from the mean position.

Answer 2

For a particle in SHM, we have

$$v^2 = \omega^2 (A^2 - y^2)$$

At the mean position $v_{\max} = \omega A$. Given:

• Amplitude $A=2\,\text{cm}=0.02\,\text{m}$
• Maximum speed $v_{\max}=10\,\text{m/s}$

So,

$$\omega = \frac{10}{0.02}=500\,\text{s}^{-1}$$

When the speed $v=5\sqrt{2}\,\text{m/s}$, apply the formula:

$$(5\sqrt{2})^2 = 500^2 \left(A^2 - y^2\right)$$ $$50 = 250000 \left(0.0004 - y^2\right)$$ $$0.0004 - y^2 = \frac{50}{250000}=0.0002$$ $$y^2 = 0.0004-0.0002 = 0.0002$$ $$y = \sqrt{0.0002}=0.01414\,\text{m} =1.414\,\text{cm}$$