First slide

Simple harmonic motion

Question

A body executes S.H.M. under the action of a force F₁ with a time period 7/6 seconds. If the force is changed to F₂ it executes S.H.M. with time period 7/8 seconds. If both the forces F₁ and F₂ act simultaneously in the same direction on the body, then its time period in seconds is

Difficult

A

7/10
B

100/49
C

1/2
D

2/1

Solution

For a body executing simple harmonic motion, the force can be written as $F = m ω^{2} y$
For the first force $F_{1} = m {(\frac{2 π}{T_{1}})}^{2} y$ and for the second force $F_{2} = m {(\frac{2 π}{T_{2}})}^{2} y$ .
On combining the two forces $F_{1} + F_{2} = F_{n e t}$ , where $F_{n e t} = m {(\frac{2 π}{T_{n e t}})}^{2} y$
Substituting this in the equation $F_{1} + F_{2} = F_{n e t}$ , we get
$m {(\frac{2 π}{T_{n e t}})}^{2} y = m {(\frac{2 π}{T_{1}})}^{2} y + m {(\frac{2 π}{T_{2}})}^{2} y$
From the question, $T_{1} = \frac{7}{6}$ s and $T_{2} = \frac{7}{8}$ s
$\begin{array}{l} {(\frac{1}{T_{n e t}})}^{2} = {(\frac{6}{7})}^{2} + {(\frac{8}{7})}^{2} \\ {(\frac{1}{T_{n e t}})}^{2} = (\frac{100}{49}) \\ T_{n e t}^{2} = (\frac{49}{100}) \end{array}$
Therefore, $T_{n e t} = (\frac{7}{10}) s$

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