First slide

Applications of SHM

Question

Due to some force F₁ a body oscillates with period 4/5 sec and due to other force F₂ oscillates with period 3/5 sec. If both forces act simultaneously, the new period will be

Moderate

A

0.72 sec
B

0.64 sec
C

0.48 sec
D

0.36 sec

Solution

Under the influence of one force $F_{1} = {mω}_{1}^{2} y$ and under the action of another force, $F_{2} = {mω}_{2}^{2} y$ .
Under the action of both the forces $F = F_{1} + F_{2}$
$\Rightarrow {mω}^{2} y = {mω}_{1}^{2} y + {mω}^{2} y$
$\Rightarrow ω_{1}^{2} + ω_{2}^{2} \Rightarrow {(\frac{2 π}{T})}^{2} = {(\frac{2 π}{T_{1}})}^{2} + {(\frac{2 π}{T_{2}})}^{2}$
$\Rightarrow T = \sqrt{\frac{T_{1}^{2} T_{2}^{2}}{T_{1}^{2} + T_{2}^{2}}} = \sqrt{\frac{{(\frac{4}{5})}^{2} {(\frac{3}{5})}^{2}}{{(\frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}}} = 0 .48 \sec$

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