Question

A body makes SHM in a line 10 cm long along y axis. Velocity at a displacement 4 cm is 12 cm/s. Time
period is
1) 3.14 sec
2) 6 28 sec
3) 2.4 sec
4) 1.57 sec​

Answer 1

Explanation:

ANSWER

v=ωA2−x2

v1=10cm/sec

v2=7cm/sec

x1=3cm

x2=4cm

v1=ωA2−x12

v2=ωA2−x22

∴710=ωA2−(4)2ωA2−(3)2

49100=A2−16A2−9

51A2=1600−(9×49)

A=

Answer 2

Answer:

The time period calculated is $1.57sec$.

Therefore, option 4) $1.57sec$ is correct.

Explanation:

Given data,

As shown in the figure;

The length of the line of the simple harmonic motion ( SHM ), 2A = $10cm$

Here, A = amplitude

The velocity at the given displacement of $4cm$, $V_{y=4}$ = $12cm/s$

The value of time period, T =?

As given,

• $2A = 10cm$
• A = $5cm$

As we know, in simple harmonic motion ( SHM ), the time period is given by:

• T = $\frac{2\pi }{\omega}$

Here, ω = angular frequency

Now, we have to calculate the value of the angular frequency by the equation given below:

• $V = \omega \sqrt{A^{2} -y^{2} }$
• $\omega = \frac{V}{ \sqrt{A^{2} -y^{2} }}$
• $\omega = \frac{12}{ \sqrt{(5)^{2} -(4)^{2} }}$
• $\omega = \frac{12}{ \sqrt{25-16}}$
• $\omega = \frac{12}{ \sqrt{9} }$
• $\omega = \frac{12}{ 3}$
• $\omega = 4rad/s$

Now, the time period is:

• T = $\frac{2\pi }{\omega}$
• T = $\frac{2\times 3.14}{4}$
• T = $\frac{6.28}{4}$
• T = $1.57s$

Hence, the time period calculated is $1.57sec$.

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