First slide

Applications of SHM

Question

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.

Moderate

A

5
B

1
C

2
D

3

Solution

If t is the time taken by pendulums to come in same phase again first time after $t = 0$ .
and $N_{S} =$ Number of oscillations made by shorter length pendulum with time period $T_{S}$ .
$N_{L} =$ Number of oscillations made by longer length pendulum with time period $T_{L}$ .
Then $t = N_{S} T_{S} = N_{L} T_{L}$
$\Rightarrow N_{S} 2 π \sqrt{\frac{5}{g}} = N_{L} \times 2 π \sqrt{\frac{20}{g}}$ ( $∵ T = 2 π \sqrt{\frac{l}{g}}$ )
$\Rightarrow N_{S} = 2 N_{L}$ i.e. if $N_{L} = 1 \Rightarrow N_{S} = 2$

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