To find the velocity of a mass attached to the midpoint of a spring

• gnits
In summary, Mitch misread the answer in the book and thought it said v^2 = 5ga. Thanks for catching that!
gnits
Homework Statement
To find the velocity of a mass attached to the midpoint of a spring
Relevant Equations
E=Yx^2/2a
Could I please ask for help regarding part (a) of this question:

If I get part (a) then part (b) will follow. So, here's my answer to part (a):

I'll be using the formula Elastic Potential Energy in a spring = (Yx^2)/(2a)
Where Y is the modulus of the spring, x the extension and a the natural length of the spring.

And we are given that Y = mg and so we have Elastic Potential Energy in this spring = (mgx^2)/(2a)

I will logically split the spring into two parts, that to the left of the particle and that to the right. Each of these has a natural length of a.

Here's a sketch of the situation.

<------------4a------------->
A --------------------------- B

We are told that the particle is pulled back until it is at point A.

So here the left hand part of the spring as an extension (compression) of length a (because natural length is a and actual length is 0)
And the right hand part an extension of 3a (because natural length is a and actual length is 4a)

So in this position the energy in the system is:

mga^2/2a + mg(3a)^2/2a

Now, when the particle is 1/4 of the way from A to B then the elastic potential energy in the left hand part of the spring is 0 because it is at it's natural length of a.

And the right hand part now has an extension of 2a (because it has length 3a and natural length a).

And the mass also has kinetic energy. Equating energy before with energy 'now' we get:

mga^2/2a + mg(3a)^2/2a = mg(2a)^2/2a + (1/2)mv^2

which leads to v^2 = 6ga

But answer in book is v^2 = 5ga

Thanks for any help in pointing out my error,
Mitch.

gnits said:
Homework Statement: To find the velocity of a mass attached to the midpoint of a spring
Homework Equations: E=Yx^2/2a

Could I please ask for help regarding part (a) of this question:

View attachment 252373

If I get part (a) then part (b) will follow. So, here's my answer to part (a):

I'll be using the formula Elastic Potential Energy in a spring = (Yx^2)/(2a)
Where Y is the modulus of the spring, x the extension and a the natural length of the spring.

And we are given that Y = mg and so we have Elastic Potential Energy in this spring = (mgx^2)/(2a)

I will logically split the spring into two parts, that to the left of the particle and that to the right. Each of these has a natural length of a.

Here's a sketch of the situation.

<------------4a------------->
A --------------------------- B

We are told that the particle is pulled back until it is at point A.

So here the left hand part of the spring as an extension (compression) of length a (because natural length is a and actual length is 0)
And the right hand part an extension of 3a (because natural length is a and actual length is 4a)

So in this position the energy in the system is:

mga^2/2a + mg(3a)^2/2a

Now, when the particle is 1/4 of the way from A to B then the elastic potential energy in the left hand part of the spring is 0 because it is at it's natural length of a.

And the right hand part now has an extension of 2a (because it has length 3a and natural length a).

And the mass also has kinetic energy. Equating energy before with energy 'now' we get:

mga^2/2a + mg(3a)^2/2a = mg(2a)^2/2a + (1/2)mv^2

which leads to v^2 = 6ga

But answer in book is v^2 = 5ga

Thanks for any help in pointing out my error,
Mitch.

Oops! It says string not spring. I'll do it again!

Yes, that solved it. Can't believe I misread that for so long.

1. What is the equation for calculating the velocity of a mass attached to the midpoint of a spring?

The equation for calculating the velocity of a mass attached to the midpoint of a spring is v = ωAcos(ωt + φ), where v is the velocity, ω is the angular frequency, A is the amplitude, t is time, and φ is the phase angle.

2. How does the mass of the object affect its velocity on the spring?

The mass of the object does not affect its velocity on the spring. The velocity is determined by the characteristics of the spring, such as its stiffness and the amplitude of its oscillations.

3. What is the relationship between the velocity and the displacement of the mass on the spring?

The velocity of the mass on the spring is directly proportional to its displacement from the equilibrium position. This means that as the displacement increases, the velocity also increases.

4. Can the velocity of the mass on the spring change over time?

Yes, the velocity of the mass on the spring can change over time. This is because the mass is constantly experiencing acceleration due to the oscillations of the spring, causing its velocity to change.

5. How can the velocity of the mass on the spring be measured experimentally?

The velocity of the mass on the spring can be measured experimentally by using a motion sensor or a high-speed camera to track its movement over time. The data collected can then be used to calculate the velocity using the appropriate equation.

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