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

2 Answers

5 votes

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

User Ricoms
by
7.0k points
7 votes

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.

User Jonauz
by
6.4k points