Question

Obtain an equation for the SHM of a particle of amplitude 0.5 m, frequency 50 Hz. The initial phase is $frac{pi}{2}$ Find the displacement at $t=0$

Answer 1

Amplitude = A = 0.5 meters
frequency = f = 50 Hz
angular frequency = ω = 2 π f = 100 π  rad/sec
initial phase = Ф = π/2

x(t) =  A Sin (ω t + Ф)
       =  0.5 Sin (100π t + π/2)
       = 0.5 Cos (100 π t)

displacement at  t = 0, 
 x(0) = 0.5 Cos 0 = 0.5 meters.

Answer 2

Answer:

The equation for the SHM of a particle of amplitude 0.5m, frequency 50 Hz. and the initial phase $\pi /2$ is    $x(t)=0.5 sin(314t+\pi /2)$

The displacement at time  $t=0$ is $0.5m$

Explanation:

• The position of a particle at any time 't' in simple harmonic motion is given by the formula

        $x(t)=A\sin \left(\omega t+\phi \right)$

        where

       $A$- the amplitude of the oscillation

       ω- angular frequency

       Φ- the initial phase

• The time period of oscillation is

        $T=\frac{2\pi }{w}$

• The frequency of oscillation

      $f=1/T$

From the question, we have

the amplitude of the oscillation $A=0.5m$

the frequency $f=50Hz$

the angular frequency ω$=2\pi f=2\pi *50=314 rad/sec$

the initial phase Ф$=\pi /2$

as compared with the equation of SHM, we can write

$x(t)=0.5 sin(314t+\pi /2)$

at time $t=0$ the displacement is

$x(0)=0.5 sin(\pi /2)=0.5m$