Question

A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. Assuming that the mass is at the equilibrium posiiton at t = 0, what is its displacement at t = 1.0 s?

Answer 1

Answer:

$d =3.7*10^{-3} m$

Explanation:

From the question we are told that:

Mass $m=1.2kg$

Spring constant $\mu=22Nm^{-1}$

Amplitude $A=5cm=0.05m$

Generally the equation for displacement d is mathematically given by

 $d = Asin(\omega t)$

Where

 $\omega=angular\ velocity$

 $\omega=\sqrt{k/m}$

 $\omega=\sqrt{22/1.2}$

 $\omega=4.2817rads^{-1}$

Therefore  

 $d = 0.05*sin(4.2817*1)$  

 $d =3.7*10^{-3} m$