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Question

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 πk
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C
The particle has angular frequency of k
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D
The particle has a time period of 2πk
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Solution

The correct option is B The particle has a time period of πk
Given,
d2vdt2=kv
ddt(dvdt)=kdxdt
dadt=kdxdt
da=kdx
Integrating both sides, we get,
a0da=kx0dx
[a=0 at x=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=2πk
Thus, option (b) is the required answer.

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