Question

a body of 0.25 kg executes SHM given by
$y = 0.4 \sin(0.5\pi \: t) m$
calculate the amplitude, angular frequency, maximum velocity and maximum acceleration..​

Answer 1

Answer:it's true

Explanation: it'strue u have to solve it.

Answer 2

Answer:

For the SHM $y=0.4sin(0.5\pi t)m$

1. amplitude $A=0.4m$
2. angular frequency ω=$0.5\pi rad/sec$
3. maximum velocity  $V_{max} =0.628m/s$
4. maximum acceleration $a_{max} =0.98m/s^{2}$

Explanation:

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

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

         where

        $A$-amplitude

       ω-angular frequency

       Ф-initial phase

• 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

                 $a_{max} =Aw^{2}$

From the question, we have

The displacement of the SHM is

$y=0.4sin(0.5\pi t)m$

as compared to the standard equation

the amplitude $A=0.4m$

the angular frequency ω=$0.5\pi rad/sec$

the maximum velocity  $V_{max} =Aw=0.4*0.5\pi =0.628m/s$

the maximum acceleration $a_{max} =Aw^{2}=0.4*(0.5\pi )^{2} =0.98m/s^{2}$