The maximum velocity of a particle performing S.H.M. is ' V '. If the periodic time is made ( 3 1 ) rt and the amplitude is doubled, then the new maximum velocity of the particle will be

Given T ′ = 3 1 T and A ′ = 2 A ∴ ω ′ = T ′ 2 π = ( 3 1 T ) 2 π = T 6 π = 3 ω ∴ The new maximum velocity V ′ = A ′ ω ′ = ( 2 A ) × ( 3 ω ) = 6 A ω = 6 V

Start Free Mock LoginAlready registered? Login here

Physics - SHM Question with Solution | TestHub

PhysicsSHMMiscellaneousMedium2 minPYQ_2023

PhysicsMediumsingle choice

The maximum velocity of a particle performing S.H.M. is ' $V$ '. If the periodic time is made $(\frac{1}{3})^{rt}$ and the amplitude is doubled, then the new maximum velocity of the particle will be

Options:

Answer:

Solution:

Given $T^{'} = \frac{1}{3} T$ and $A^{'} = 2 A$ $∴ ω^{'} = \frac{2 π}{T ^{'}} = \frac{2 π}{( \frac{1}{3} T )} = \frac{6 π}{T} = 3 ω$ $∴$ The new maximum velocity $V^{'} = A^{'} ω^{'} = (2 A) \times (3 ω) = 6 A ω = 6 V$

Stream:MHT_CETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2023

Doubts & Discussion

Loading discussions...