Question

A mass is oscillating on a spring with amplitude 8.3. m
and a period of 8.5 s. What is the maximum velocity of
the mass?​

Answer 1

Since the question deals with a particular case of maximum velocity it makes such question a real easy one if you know the theory and formula part well. So, here's all the basic thing you need to know to solve such type of question :

If a particle of mass m is oscillating with an amplitude A,time period T and frequency n then the velocity of the particle in terms of displacement x is given by :

• V = ω √(A² -x²)

Now,if you want to have maximum velocity, you put x = 0 in the formula to obtain Vmax.

• Vmax = ωA

And if you want minimum velocity, you put x = A in the velocity formula,

• Vmin = ω√(A² - A²) = ω(0) = 0

So from above result we conclude the following :

Velocity of a particle performing oscillation is maximum at the mean position. Whereas, it is minimum at the extreme position.

Now moving towards the question.

Given data :

The amplitude of the oscillation,A = 8.5 m and the period of oscillation,T = 8.5 s.

Solution :

We are well aware of the fact that the frequency,n = 1/T ----> (1)

And we now know the formula for maximum velocity which is Aω.

So,if you look at the formula for max velocity you will find something is missing from this question. We don't yet have the value of ω,so first we will find ω.

We know time period,T = 8.5 s so substitute this value in (1),

n = 1/8.5

n = 0.11 ----> (2)

Hope you know that ω = 2πn so using this we have :

ω = 2 × π × 0.11

ω = 0.70 ----> (3)

Now just substitute this value from (3) in the formula of Vmax.

Vmax = ωA

Vmax = (0.70) (8.3)

Vmax = 5.81 m/s

So,the maximum velocity of the mass is 5.81 m/s.