Question

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

Short answer

a. Oscillation frequency is$3.18\mathrm{Hz}$.

b. Amplitude of the motion is $7.07\mathrm{cm}$.

c. Total mechanical energy is$5\mathrm{J}$.

Step-by-step solution

Calculation for frequency (part a)

a.

Given info:

The mass is $5\mathrm{Kg}$.

The spring constant is $2000\mathrm{N}/\mathrm{m}$.

Time period is,

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{5\mathrm{kg}}{2000{\displaystyle \frac{\mathrm{N}}{\mathrm{m}}}}}$

$=0.314\mathrm{s}$

Oscillation frequency is defined as number of oscillation per second.

So,

$\mathrm{f}=\frac{1}{\mathrm{T}}$

$=\frac{1}{0.314\mathrm{s}}$

$=3.18\mathrm{Hz}$

Calculation for total mechanical energy (part c)

c.

Total mechanical energy is,

$\mathrm{E}=\mathrm{K}+\mathrm{U}$

$=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{kx}}^2$

$=\frac{1}{2}(5\mathrm{kg})(1\mathrm{m}/\mathrm{s}{)}^2+\frac{1}{2}(2000\mathrm{N}/\mathrm{m}\left)\right(0.05\mathrm{m}{)}^2$

$=5\mathrm{J}$

Calculation for Amplitude (part b)

b.

Total mechanical energy is related to amplitude is,

$\mathrm{E}=\frac{1}{2}{\mathrm{kA}}^2$

$\mathrm{A}=\sqrt{\frac{2\mathrm{E}}{\mathrm{K}}}$

$\mathrm{A}=\sqrt{\frac{2\times 5\mathrm{J}}{2000{\displaystyle \frac{\mathrm{N}}{\mathrm{m}}}}}$

$\mathrm{A}=\sqrt{\frac{2\times 5\mathrm{J}}{2000{\displaystyle \frac{\mathrm{N}}{\mathrm{m}}}}}$

role="math" localid="1650019534332" $\mathrm{A}=7.07\mathrm{cm}$