Question

The displacement of a particle performing linear S.HM, is given by x6 sin 3
$\pi$
t+5
$\pi$
/6mPind amplitude,
frequency and the phase constant of the motion​

Answer 1

Answer:

solved

Explanation:

s = 6 sin (3 t+5)

amplitude = 6

frequency = 3/2π

phase constant of the motion​ = 5

Answer 2

Answer:

For the simple harmonic motion, $x(t)=6sin(3\pi t+5\pi /6)$ the phase constant is $(3\pi t+5\pi /6)$ and the frequency of oscillation $f=1.5Hz$

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

       $\left(\omega t+\phi \right)$- phase constant

• The frequency of oscillation is

        $f={w}/2\pi$

From the question, we have

$x(t)=6sin(3\pi t+5\pi /6)$

as compared with the SHM equation, we get

the phase constant of the motion is

$(3\pi t+5\pi /6)$

the angular frequency ω$=3\pi$

and the frequency of oscillation $f=w/2\pi =3\pi /2\pi =3/2=1.5Hz$