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, $\nu=\dfrac{1}{8}\ Hz$

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 :

$y=5\ sin(\dfrac{\pi}{4})\ t$............(2)

Comparing equation (1) and (2)

$\because\ \omega=2\pi\nu$

So,

$\dfrac{\pi}{4}=2\pi \nu$

$\nu=\dfrac{1}{8}\ Hz$

Hence, this is the required solution.