Question

A particle is vibrating in simple harmonic motion of amplitude 20 cm and time period 0.5s. Find its maximum speed and speed when its displacement is 10cm. ​

Answer 1

Answer:

• Maximum Velocity (v) = 0.2512 m/s.
• Speed at Displacement 10 cm is (v) = 0.2175 m/s.

Given:

1. Amplitude of Particle (A) = 20 cm.
2. Time period (t) = 0.5 seconds.
3. Displacement (y) = 10 cm.

Explanation:

$\rule{300}{1.5}$

From the Formula we Know,

$\large\bigstar \: {\boxed{\tt v_{max} = \omega A}}$

$\bold{Here}\begin{cases}\text{A Denotes Amplitude} \\ \omega \text{ Denotes Angular Frequency}\end{cases}$

Now,

$\large{\boxed{\tt v_{max} = \omega A}}$

$\large{\tt \longmapsto v_{max} = \Bigg(\dfrac{2 \pi}{T} \Bigg) \times A}$

Substituting the values,

$\large{\tt \longmapsto v_{max} = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 20 \: cm}$

As Amplitude is given in Cm we need to convert it into Meters.

∵ [1 cm = 10⁻³ m]

$\large{\tt \longmapsto v_{max} = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 20 \times 10^{-3} \: m}$

$\large{\tt \longmapsto v_{max} = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 20 \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = \Bigg(\dfrac{2 \times 10 \pi}{5} \Bigg) \times 20 \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = \Bigg(\cancel{\dfrac{2 \times 10 \pi}{5}} \Bigg) \times 20 \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = \Bigg(\dfrac{2 \times 2 \pi}{1} \Bigg) \times 20 \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = 2 \times 2 \times \pi \times 20 \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = 80 \times \pi \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = 80 \times 3.14 \times 10^{-3}}$

$\large{\tt \longmapsto v_{max} = 251.2 \times 10^{-3}}$

$\large\longmapsto{\underline{\boxed{\red{\tt v_{max} = 0.2512 \; m/s}}}}$

∴ Maximum Velocity (v) = 0.2512 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Formula we Know,

$\large\bigstar \: {\boxed{\tt v = \omega \sqrt{A^2 - y^2}}}$

$\bold{Here}\begin{cases}\text{y Denotes Displacement} \\ \omega \text{ Denotes Angular Frequency}\\ \text{A Denotes Amplitude}\end{cases}$

Now,

$\large{\boxed{\tt v = \omega \sqrt{A^2 - y^2}}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{T} \Bigg) \times \sqrt{A^2 - y^2}}$

Substituting the values,

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times \sqrt{(20)^2 - (10)^2}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times \sqrt{400 - 100}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times \sqrt{300}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 10 \sqrt{3}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 10 \sqrt{3} \: cm}$

As it is given in Cm we need to convert it into Meters.

∵ [1 cm = 10⁻³ m]

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 10 \sqrt{3} \times 10^{-3} \: m}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \pi}{0.5} \Bigg) \times 10 \sqrt{3} \times 10^{-3}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \times 10 \pi}{5} \Bigg) \times 10 \sqrt{3} \times 10^{-3}}$

$\large{\tt \longmapsto v = \Bigg(\cancel{\dfrac{2 \times 10 \pi}{5}} \Bigg) \times 10 \sqrt{3} \times 10^{-3}}$

$\large{\tt \longmapsto v = \Bigg(\dfrac{2 \times 2 \pi}{1} \Bigg) \times 10 \sqrt{3} \times 10^{-3}}$

$\large{\tt \longmapsto v = 2 \times 2 \times \pi \times 10 \sqrt{3} \times 10^{-3}}$

$\large{\tt \longmapsto v = 40 \sqrt{3} \times \pi \times 10^{-3}}$

$\large{\tt \longmapsto v = 40 \sqrt{3} \times 3.14 \times 10^{-3}}$

$\large{\tt \longmapsto v = 125.6 \sqrt{3} \times 10^{-3}}$

$\large{\tt \longmapsto v = 125.6 \times 1.732 \times 10^{-3}}$

$\large{\tt \longmapsto v = 217.5 \times 10^{-3}}$

$\large\longmapsto{\underline{\boxed{\red{\tt v_{max} = 0.2175 \; m/s}}}}$

∴ Speed at Displacement 10 cm is (v) = 0.2175 m/s.

$\rule{300}{1.5}$

Answer 2

A very important property of simple harmonic motion is that the period T does not depend on the amplitude of the motion, A.