Test Index

Oscillations Part 1

Section: Physics

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Question : 30 of 100

Marks: +1, -0

The point

A

moves with a uniform speed along the circumference of a circle of radius

0.36\ \mathrm{m}

30^{\circ}

0.1\ \mathrm{s}

. The perpendicular projection

P

A

on the diameter

M N

represents the simple harmonic motion of

P

. The restoration force per unit mass when

P

M

[25 Feb 2021 Shift 2]

100 N
9.87 N
50 N
0.49 N

Solution:

Given, radius of circle,

R=0.36\ \mathrm{m}

Angular distance,

\theta=30^{\circ}=\frac{\pi}{6}\ \mathrm{rad}

Let

/

be the arc length.

\because \;\; I=R\theta

\Rightarrow \;\; I=\;\frac{36}{100} \times \;\frac{\pi}{6}=\;\frac{6\pi}{100}\ \mathrm{m}

A s, speed on circular track

(v)=\text{Arclength}(l)/\text{Time}(t)

\Rightarrow \;\; v=\;\frac{6\pi}{100 \times 0.1}=\;\frac{6\pi}{10}\ \mathrm{ms}^{-1}

If

F

be the restoration force and

a_{t}

be the radial acceleration

(=v^{2}/R)

, then

F \;\; =m a_{r}

a_{r} \;\; =\;\frac{F}{m}=\;\frac{v^{2}}{R}=\left(\;\frac{6\pi}{10}\right)^{2} \times \;\frac{100}{36}

\;\; =\;\frac{36 \times 9.87}{100} \times \;\frac{100}{36}=9.87\ \mathrm{N}

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