# Block on a vertical spring, finding frequency

1. Mar 28, 2015

### laurenm02

1. The problem statement, all variables and given/known data
A block of mass m is on a platform of mass M, supported by a vertical, massless spring with spring constant k.

When the system is at rest, how much is the spring compressed?
When the spring is pushed an extra distance x, what is the frequency of vertical oscillation?

2. Relevant equations
F = -kx

3. The attempt at a solution
For the first question, I set F = mg = -kx, where m = (M + m), and found x = (M+m)g/-k

For the second part, I'm not sure how to set it up. I first thought that frequency is independent of amplitude, but I was told that I have to first write a new second law equation and then find an equation of motion.

2. Mar 28, 2015

### sk1105

Frequency will not be independent of amplitude, because the frequency depends on the force exerted by the spring, which in turn depends on the amplitude of the compression. So yes, you'll have to write out a new second law equation taking into account the extra compression.

3. Mar 28, 2015

### laurenm02

Can I get a little guidance on this? I'm not sure how to set it up.

4. Mar 28, 2015

### I like Serena

Hi laurenm02! :)

Sorry, but frequency is independent of amplitude.

The mass M should not be included. It does not matter whether the spring rests on a platform of mass M or directly on the ground.
What matters, is that we have a mass m that is supported by the spring.
So the compression at rest should be $\frac{mg}{k}$.

Now suppose we compress the spring by an additional amount $x$.
Then the resultant force becomes $F_{result} = k(x_0 - x) - mg$.
Newton's second law states that $F_{result} = ma$.
So we get that:
$$ma = k(x_0 - x) - mg$$
Or, written as a differential equation:
$$m\frac{d^2x}{dt^2} = k(x_0 - x) - mg$$

Are you supposed to be able to solve such an equation?

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