The velocity of a particle of mass m moving along a straight line varies with time t as d2vdt2=−kv, the particle's acceleration was zero when it started its motion from x=0. Here k is a positive constant. Choose the incorrect statement.
A
The particle is performing SHM
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B
The particle has a time period of π√k
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C
The particle has angular frequency of √k
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D
The particle has a time period of 2π√k
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Solution
The correct option is B The particle has a time period of π√k Given, d2vdt2=−kv ⇒ddt(dvdt)=−kdxdt ⇒dadt=−kdxdt ⇒da=−kdx
Integrating both sides, we get, ∫a0da=−k∫x0dx [∵a=0atx=0] ⇒a+kx=0 ⇒d2xdt2+kx=0
This equation is similar to the equation of free oscillations. ∴ Particle performs SHM.
On comparing with the equation , d2xdt2+ω2ox=0, we get,
Angular frequency, ωo=√k
Now, Time period of oscillation, T=2πωo=2π√k
Thus, option (b) is the required answer.