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Rudik [331]
4 years ago
15

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

Physics
2 answers:
lord [1]4 years ago
8 0

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

andrew11 [14]4 years ago
8 0

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.

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