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 .

1. Determine the speed when it passes the equilibrium point.

2. Determine the speed when it is 0.12 m from equilibrium.

3. Determine the total energy of the system.

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

Answer 2

1) The speed of the mass at the end of a spring is = 2.37 m/s

2) The speed of the mass when it is 0.12 m from the equilibrium = 0.91 m/s

3) The Total energy of the system = 0.62 J

Given data :

Mass at the end of the spring = 0.22 kg

Frequency = 2.9 Hz

Amplitude of mass = 0.13 m

k = spring constant

1) Determine the speed of the mass as it passes through the equilibrium point

First step : Determine the Angular velocity

ω = 2*π*f

   = 2 * π * 2.9 = 18.22 rad/s

Next step : Determine the speed/velocity as it passes through the equilibrium point

V = A * ω

  = 0.13 * 18.22 = 2.3686 m/s ≈ 2.37 m/s

2) Determine the speed when it is 0.12 m from equilibrium

First step : calculate the value of K

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

k = m*4π² * f²

k = 0.22 * 4π² * 2.9²

  = 73.04 N / m

displacement from equilibrium ( x ) = 0.12 m

Note : Given that Energy is conserved in the system

Final step :  determine Speed when mass is displaced 0.12 m from the equilibrium position

speed ( V ) = $\sqrt{\frac{k(A^2-x^{2} )}{m} }$  ------ ( 1 )

where ; A = 0.13,   K = 73.04,  x = 0.12,  m = 0.22

insert values into equation ( 1 ) above

∴ Speed ( V ) ≈  0.91 m/s

3)  Calculate the total energy of the system

Total energy = $\frac{1}{2} KA^{2}$

E = 1/2 * 73.04 * ( 0.13 )²

  = 0.617 ≈ 0.62 J

Hence we can conclude that; The speed of the mass at the end of a spring is = 2.37 m/s, The speed of the mass when it is 0.12 m from the equilibrium = 0.91 m/s, The Total energy of the system = 0.62 J .

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