Question

A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium. What are the (a) frequency, (b) amplitude, and (c) total mechanical energy of the motion?

Answer 1

Explanation:

Given that,

Mass of the block, m = 5 kg

Spring constant, k = 2000 N/m

The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium.

(a) The angular frequency is given by :

$\omega=\sqrt{\dfrac{k}{m}}$

Since, $\omega=2\pi f$, f is frequency

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{2000}{5}} \\\\f=3.18\ Hz$

(b) The velocity of particle executing SHM is given by :

$v=\omega\sqrt{A^2-x^2}$

x is displacement from equilibrium position

$\dfrac{v}{\omega}=\sqrt{A^2-x^2} \\\\(\dfrac{v}{\omega})^2=A^2-x^2\\\\A^2=(\dfrac{v}{\omega})^2+x^2\\\\A=\sqrt{(\dfrac{v}{\omega})^2+x^2} \\\\A=\sqrt{(\dfrac{1}{2\pi \times 3.18})^2+(0.05)^2}\\A=0.070\ m\\\\A=7\ cm$

(c) The total mechanical energy of the motion in SHM is given by :

$E=\dfrac{1}{2}kA^2\\\\E=\dfrac{1}{2}\times 2000\times (0.070)^2\\\\E=4.9\ J$

Hence, this is the required solution.