Question

Plz Answer it urgently ........
If the maximum velocity and the acceleration of a particle executing SHM are equal in magnitude then the time period is equal to :
A) $\pi$ sec
B) $\frac{ \pi }{2}$ sec
C) 2$\pi$ sec
D) $\frac{2}{ \pi }$ sec

Answer 1

Answer is (C) 2π sec.

SHM:

Let the displacement of the particle be x, the angular frequency be ω , and the amplitude be A.  Let v and a be the velocity and acceleration respectively.  Then at time t,

     x = A Sin ωt
     v = A ω Cos ωt
     a = - A ω² Sin ωt = - ω² x

Given:   Maximum velocity = A ω = Maximum Acceleration = A ω²

 =>  ω = 1 rad/sec
=>   T  = 2π/ω 
            = 2π Sec.

Answer 2

Answer:

If the magnitude of maximum velocity is equal to the maximum acceleration of a particle executing SHM then the time period is equal to $2\pi$ sec

Explanation:

• In simple harmonic motion, the displacement of a particle is given by the formula

                  $x(t)=Asin(\omega t+\phi )$

• The velocity of that particle is

                 $dx(t)/dt=Awsin(\omega t+\phi )$

         we get maximum velocity when sine is maximum, that is 1

                  $V_{max} =Aw$

• The acceleration of that particle is

                 $d^{2} x(t)/dt^{2} =Aw^{2} sin(\omega t+\phi )$

         maximum acceleration is

                 $a_{max} =Aw^{2}$

• For a particle in SHM with angular frequency 'ω', the time for one oscillation or the time period is

                 $T=2\pi /w$

From the question, we have

maximum acceleration = maximum velocity

⇒

$V_{max} =a_{max} \\Aw=Aw^{2}\\w=1$

substitute the value of ω in time period equation, we get

$T=2\pi /w=2\pi /1=2\pi sec$