Question

A mass of 0.40 kg, hanging from a spring with a spring constant of 80 N/m, is set into an up-and-down simple harmonic motion. What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.

Answer 1

Answer:

v = 1.22 m/s

Explanation:

Given that,

Mass of the object, m = 0.4 kg

Spring constant of the spring, k = 80 N/m

The starting displacement from equilibrium is 0.10 m, or the amplitude, A = 0.1 m

To find,

The speed of the mass when moving through the equilibrium point.

Displacement in the mass, y = 0.05

Solution,

Let $\omega$ is the angular frequency of the mass. It can be calculated as :

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

$\omega=\sqrt{\dfrac{80}{0.4}}$

$\omega=14.14\ rad/s$

When the mass displaces from some position, the speed of the mass is given by :

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

$v=14.14\sqrt{(0.1)^2-(0.05)^2}$

v = 1.22 m/s

So, the speed of the mass when moving through the equilibrium point is 1.22 m/s.

Answer 2

The speed of the mass when moving through the equilibrium point is 1.22 m/s.

Calculation of the mass speed:

But before that the angular frequency is

$= \sqrt{\frac{80}{0.4} }$

= 14.14 rad/s

Now

The mass speed should be

$= 14.14 \sqrt{(0.10)^2 - (0.05)^2}$

= 1.22 m/s

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