Harmonic Motion of Pendulum on Mars

[rzermatt]

[SOLVED] Harmonic Motion of Pendulum on Mars

Homework Statement

The problem is as follows: The acceleration due to gravity on Mars in 3.7m/s^2.

A mass is suspended from a spring with a force 10 N/m. Find the mass suspended from this spring that would result in a one second period on Earth and on mars.

Homework Equations

Obviously something has to be done with:
T = 2\pi\sqrt{}m/k

The Attempt at a Solution

I have no idea where to start with this problem. I know that T is going to equal one for both. I also know that there is definitely a direct proportionality between mass and stretch constant as well as stretch. Where does gravity come into play here? Would the gravity matter? When I visualize the problem it does, so how could I apply this here? Or am I overthinking?

Thanks in advance, and I will reply as quickly as possible to any replies.

 

[Gear300]

Yup, that is the equation you use...but if you noticed, its almost a trick. Notice that gravity is nowhere in the equation, therefore, the masses would be the same for Earth and Mars. When suspending a spring vertically, gravity stretches it down a bit, but what happens is that the equilibrium point simply shifts to a new position; otherwise, gravity does not play much of a part in the oscillations.

 

[rzermatt]

Thanks, that makes perfect sense to me. Now that I think about it, the equilibrium point will change but the change of the mass on the period will not since the spring constant remains the same at both locations.

 

[Ithryndil]

This may be a dumb question, but would gravity simple play a role in the amplitude of a spring for any given period. What I mean is this. If let's say we want the period to be 1, then wouldn't the amplitude of that mass on Earth be greater than the amplitude of that mass on Mars. And wouldn't the amplitude for Mars be greater than an amplitude for the moon for the same period of 1 second?

 

[rzermatt]

This may be a dumb question, but would gravity simple play a role in the amplitude of a spring for any given period. What I mean is this. If let's say we want the period to be 1, then wouldn't the amplitude of that mass on Earth be greater than the amplitude of that mass on Mars. And wouldn't the amplitude for Mars be greater than an amplitude for the moon for the same period of 1 second?

That would make sense to me, and of course it would agree with what was previously stated. Gravity on any of those planets would merely affect the amplitude which would not change the period of the oscillation and not change the mass requirements according to the problem.

 

[Gear300]

There is no dumb question...or at least this one is not. The amplitude simply depends on how much energy you put into the system, which would be 1/2*k*x^2...gravity does not affect it. Gravity is a conservative force; its not going to change the energy of the system.

 

[Ithryndil]

That was my thought. Because the gravity is greater on Earth, the equilibrium position would be shifted and the amplitude would be greater as well, of course, because the amplitude would be greater, vmax for the mass on the Earth would be greater than vmax for the mass on Mars.

 

[Ithryndil]

Hmm. Well, it still seems to me that one could argue that if the spring is vertical, then gravity will play a role in the gravitational potential energy of the system, and if g is larger in one place than another, then that potential energy would be greater (assuming the same height in all cases).

 

[Gear300]

Since it shifts the equilibrium point, then yes, you're right. But since one can negate its effect during the oscillations, I'm thinking that the amplitudes would be unaffected.

 

[Ithryndil]

Hmm. Perhaps you're right. I am just thinking that the greater the gravity is, the greater the amplitudes, but I don't think there is an equation relating gravity and amplitudes. My reason was that if g is larger at x than at y, then x would have a greater A, but that would require it to have a larger vmax to satisfy the same period.