Question

The speed of a 2 kg mass on a spring is 5 m/s as it passes through its equilibrium position. What is its frequency if the amplitude is 2.5 m

Answer 1

Solution :

Given :

Mass of the object attached to the spring, m = 2 kg

Velocity of the object as it moves, V = 5 m/s

Amplitude of the object as it swings, A = 2.5 m

We have to find the frequency of the object.

We known,

$V_{max} = \omega A = 2 \pi f A$

Therefore, $f= \frac{V_{max}}{2 \pi A}$

$f= \frac{5}{2 \times 3.14 \times 2.5}$

f   = 0.32 Hz

Therefore the frequency of the object is 0.32 hertz when the amplitude of the 2 kg mass is 2.5 m moving a speed of 5 m/s.