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

Answer:

Time taken by the object is 0.012 s .

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 , $x = 4.00\times 10^{-2}\ m.$

Putting all these value in equation 1 we get ,

$4\times 10^{-2}=0.15 \times sin(22\times t) \\\\0.27=sin(22\times t)\\\\22\times t=sin^{-1}{0.27}\\\\t=\dfrac{sin^{-1}0.27}{22}\\\\t=0.012\ s .$

Hence , this is the required solution.