Question

The maximum speed of a particle performing liner shm is 0.08m/s.if maximum accleleration is 0.32m/S2 calculate it's period and amplitude.

Answer 1

Answer:

The maximum speed of a particle performing liner SHM is 0.08m/s, if maximum acceleration is 0.32m/S2, the time period is $1.57s$ and the amplitude of oscillation is $0.02m$

Explanation:

• In simple harmonic motion restoring force is directly proportional to the acceleration of the body

• The displacement of the body executing SHM is given by the formula

                       $x(t)=A sin(wt+$Ф)

         where,

        $A$-amplitude of the oscillation

        ω-the angular frequency

        Ф-phase

• The speed of the particle is $v(t)=dx/dt=A wsin(wt+$Ф), and the maximum speed is when the sine value is one $V_{max}=Aw$

• The acceleration of the particle $a=dv/dt=Aw^{2} sin(wt+$Ф), the maximum acceleration is $a_{max}=Aw^{2}$

• The time period is the time taken to complete one oscillation, it is given by the formula $T=2\pi /w$

From the question, we have

maximum velocity is $Aw=0.08m/s$ (equation 1)

maximum acceleration is $Aw^{2} =0.32m/s^{2}$(equation 2)

equation2/equation 1

⇒$\frac{Aw^{2} }{Aw } =w=0.32/0.08=4rad/sec$

The time period $T=2\pi/4=1.57s$

we have $Aw=0.08m/s$  by substituting the value of $w$ we will get amplitude as

$A=0.08/4=0.02m$