Start Free Mock LoginAlready registered? Login here

Physics - SHM Question with Solution | TestHub

PhysicsSHMKinematicsEasy2 minQB

PhysicsEasysingle choice

Two simple pendulums of length $5 m$ and $20 m$ are given small displacements in the same direction at the same time. The minimum number of oscillations, the shorter pendulum has completed, when the phase difference between them becomes zero again, is

Options:

Answer:

Solution:

$N_{S}$ = 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 $T_{L}$ .

If $t$ is the time after which the pendulums are back in sync, then

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 $\Rightarrow N_{L} = 1$ $N_{S} = 2$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Kinematics

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...