Question

An object is undergoing simple harmonic motion with frequency f = 3.5 Hz and an amplitude of 0.15 m. At t = 0.00 s the object is at x = 0.00 m. How long does it take the object to go from x = 0.00 m to x = 4.00×10−2 m.

Answer 1

Time taken by the object is 0.012 s .

Step-by-step explanation:

Given :

Frequency , f = 3.5 Hz .

Amplitude , A = 0.15 m .

At time t = 0 , x = 0 m.

Since , at time t = 0 , x = 0 m .

Therefore , equation of displacement is :

`x=Asin(\omega t)` ...equation 1.

Here ,
`\omega` is angular frequency and is given by :

`\omega=2\pi f=22\ Hz.`

We need to find the time at which its displacement is ,
`x = 4.00* 10^(-2)\ m.`

Putting all these value in equation 1 we get ,

`4* 10^(-2)=0.15 * sin(22* t) \\\\0.27=sin(22* t)\\\\22* t=sin^(-1){0.27}\\\\t=(sin^(-1)0.27)/(22)\\\\t=0.012\ s .`

Hence , this is the required solution.