Solve Homework: Springs & Pendulums

[moonlit]

I have two homework problems that I'm really stuck on. Can someone please help me out? Thanks!

1) A 0.55-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstretched length triples. What is the mass of the second block?

2) A simple pendulum is made from a 0.655-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

 

[jamesrc]

1. What do you know about Hooke's Law? For the first part, m₁g = k(x-x_o), meaning the weight is balanced by the spring force. For the second part, m₂g = k(3(x-x_o)), reflecting the tripled spring extension. Since it's the same spring, the value of k does not change. Solve for m₂.

2. The pendulum has its greatest speed at its lowest point (when all of the gravitational potential energy is converted to kinetic energy). You should know a formula that relates the period of oscillation to the length of the pendulum. Use this to calculate the period, then realize that the pendulum reaches the bottom of its trajectory at t = T/4.

 

[moonlit]

I still can't get an answer for the first one. I was told to try using a formula like this:

g(m+mx)=-K3x
g(m+mx)=3(-kx)
g(m+mx)=3mg
m+mx=3m
mx=?

But I keep coming up with 7.1 kg which sounds way off. Ughh what am I doing wrong?!

 

[jamesrc]

Originally posted by moonlit
I still can't get an answer for the first one. I was told to try using a formula like this:

g(m+mx)=-K3x
g(m+mx)=3(-kx)
g(m+mx)=3mg
m+mx=3m
mx=?

But I keep coming up with 7.1 kg which sounds way off. Ughh what am I doing wrong?!

What you've written out is fine, assuming mx is a variable meaning the added mass. When you solve that last line, don't you come up with mx = 2m? Just plug in m = .55 kg and that's all there is to it. (mx = 1.1 kg)

 

[moonlit]

Ok, I understand the first problem but I'm not so sure about the second one. I used the formula:

w= sqrt g/L
w=2pi/T
T=2piw

I keep getting the answer .52 but I'm sure I'm doing something wrong on this one too. Ughhhhh!

 

[jamesrc]

It may have been a typo, but you should have:

$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{l}{g}}$

Just plug in the right numbers to find T, then remember that the answer to the question is t = T/4.

 

[moonlit]

Ok, thanks! I got the correct answer of 0.41 seconds! :)