Question

For the simple harmonic motion equation d=5sin((pi/4)t), what is the frequency?

Answer 1

it's just 1/8, no pi included (if you're on APEX)

Answer 2

Answer : Frequency,
`\\u=(1)/(8)\ Hz`

Explanation :

The equation of a particle executing SHM is given by :

`y=A\ sin\omega t`.........(1)

Where,

y is the displacement

A is the amplitude of the wave

`\omega` is the angular frequency of the wave

t is the time

The given equation is :

`y=5\ sin((\pi)/(4))\ t`............(2)

Comparing equation (1) and (2)

`\because\ \omega=2\pi\\u`

So,

`(\pi)/(4)=2\pi \\u`

`\\u=(1)/(8)\ Hz`

Hence, this is the required solution.