Question

a 500 g mass is suspended by a spring with a spring constant of 100N/m. The mass is displaced 10 cm from equilibrium and released. Determine: a) the speed of the mass as it passes through the equilibrium point. b) the speed of the object when it is 5 cm from the equilibrium position. c) the time taken by the mass to complete one cycle. d) the frequency of the oscillation e) an equation for the position of the object in terms of time. Assume t=0 when mass is released.

Answer 1

That is an harmonic motion, described by the equation

$y(t)=A\sin\left(\omega t+\phi)$

And the velocity is

$v(t)=A\omega\cos\left(\omega t+\phi)$

where

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

a) When the mass through the equilibrium point speed reach the maximum value

$v_{\rm max}=A\sqrt\frac{k}{m}$

The amplitude is the initial displacement $A=0.10m$ then the maximum speed is

$v_{\rm max}=(0.10m)\sqrt\frac{(100N/m)}{(0.5kg)}$

$v_{\rm max}=1.41m/s^2$

b) The time that takes the 0.5cm from the equilirium position is

$5cm=10cm\sin\left(\omega t+\phi)$

$\frac{1}{2}=\sin\left(\omega t+\phi\right)$

$\arcsin\left( \frac{1}{2}\right )=\omega t+\phi$

$\omega t=\arcsin\left( \frac{1}{2}\right )-\phi$

$t=\frac{1}{\omega}\left[ \arcsin\left( \frac{1}{2}\right )-\phi\right ]$

Replacing that in the equation of the velocity

$v=A\omega\cos\left(\omega \frac{1}{\omega}\left[ \arcsin\left( \frac{1}{2}\right )-\phi\right ]+\phi\right)$

$v=A\omega\cos\left(\arcsin\left( \frac{1}{2}\right )-\phi+\phi\right)$

$v=A\omega\cos\left(\arcsin\left( \frac{1}{2}\right )\right)$

$v=(0.10m)\sqrt\frac{100N/m}{0.5kg}\cos\left(\arcsin\left( \frac{1}{2}\right )\right)$

$v=1.22m/s$

c) The period in an harmonic motion is

$T=\frac{2\pi}{\omega}$

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

$T=2\pi\sqrt\frac{0.5kg}{100N/m}$

$T=0.44s$

d) The frequency is

$f=\frac{\omega}{2\pi}$

$f=\frac{1}{2\pi}\sqrt\frac{k}{m}$

$f=\frac{1}{2\pi}\sqrt\frac{100N/m}{0.5kg}$

$f=2.25Hz$

e) We already have the equation of the motion but whe dont have yet $\phi$. We can find it since we knoe that in t=0 y=10cm.

$A=A\sin\left(\omega (0)+\phi)$

$1=\sin\left(\phi)$

$\arcsin(1)=(\phi)$

$\frac{\pi}{2}=\phi$

Then reaplcing the values in the principal equation $\omega=\sqrt{(100N/m)/(0.5kg)}=10\sqrt2 s^{-1}$

$y(t)=(0.1m)\sin\left(10\sqrt2 \, s^{-1}t+\frac{\pi}{2}\right)$