# A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its time period in s.

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A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its time period in s.

(a) 1.1

(b) 1.2

(c) 1.3

(d) 1.4

The question was posed to me in my homework.

The query is from Differential Equation of Simple Harmonic Motion topic in section Simple Harmonic Motion of Machine Kinematics

by (87.2k points)

Explanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation

v = $\omega \sqrt{r^2 – x^2}$

Given values:

when point (x) = 0.75 m, then its velocity (v) = 11 m/s

when point (x) = 2 m, then its velocity (v) = 3 m/s

When point is 0.75 m away from the mid path (v) is,

v = $\omega \sqrt{r^2 – x^2}$

11 = $\omega \sqrt{r^2 – (0.75)^2}$ —>Eq(1)

Similarly, When point is 2 m away from the centre (v) is,

v = $\omega \sqrt{r^2 – x^2}$

3 = $\omega \sqrt{r^2 – 2^2}$ —>Eq(2)

Solving Eq(1) & Eq(2) we get,

$\frac{11}{3} \frac{\omega \sqrt{r^2 – (0.75)^2}}{\omega \sqrt{r^2 – 2^2}} = \frac{\sqrt{r^2 – (0.75)^2}}{\sqrt{r^2 – 2^2}}$

Squaring on both sides, we get

$\frac{121}{9} = \frac{r^2 – 0.5625}{r^2 – 4}$

121r^2 – 484 = 9r^2 – 5.0625

121r^2 – 9r^2 = 484 – 5.0625

112r^2 = 478.9375

r^2 = $\frac{478.9375}{112}$

r^2 = 4.27622

r = 2.07m

Substituting the value of r in Eq(1) we get,

11 = $\omega \sqrt{(2.07)^2 – (0.75)^2}$

11 = $\omega \sqrt{4.2849 – 0.5625}$

11 = $\omega \sqrt{3.7224}$

11 = 1.9293 ω

ω = $\frac{11}{1.9293}$ = 5.7 rad/s.

We know periodic time is,

Tp = $\frac{2 \pi}{\omega} = \frac{2 \pi}{5.7} = \frac{2 \times 3.14}{5.7} = \frac{6.28}{5.7}$ = 1.1 s.

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