Question

A 0.22 kg mass at the end of a spring oscillates 2.9 times per second with an amplitude of 0.13 m .

Answer 1

Answer:

2.3687599 m/s

0.91106 m/s

0.617213012 J

Explanation:

f = Frequency = $2.9\ Hz$

A = Amplitude = 0.13 m

k = Spring constant

m = Mass of object = 0.22 kg

Angular velocity is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi 2.9\\\Rightarrow \omega=18.22123\ rad/s$

Velocity is given by

$V=A\omega\\\Rightarrow V=0.13\times 18.22123\\\Rightarrow V=2.3687599\ m/s$

Speed when it passes the equilibrium point is 2.3687599 m/s

Frequency is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow k=m4\pi^2f^2 \\\Rightarrow k=0.22\times 4\pi^2\times 2.9^2\\\Rightarrow k=73.04296\ N/m$

x = Displacement = 0.12 m

In this system the energies are conserved

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{73.04296(0.13^2-0.12^2)}{0.22}}\\\Rightarrow v=0.91106\ m/s$

The speed when it is 0.12 m from equilibrium is 0.91106 m/s

The energy in the system is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow E=\dfrac{1}{2}\times 73.04296\times 0.13^2\\\Rightarrow E=0.617213012\ J$

The total energy of the system is 0.617213012 J