# Homework Help: Natural frequency of a spring(self-weight only)

1. Oct 27, 2007

natural frequency of a spring(self-weight only)

giving the length,mass, and spring constant of a spring...how do you find the resonance frequency..having been thinking about this question for so long, does anyone know how to approach this question...

Last edited: Oct 28, 2007
2. Oct 27, 2007

natural frequency of a spring(self-weight only)

1. The problem statement, all variables and given/known data
a 10 cm long brass coil with spring constant 2 N/m, mass 0.1 kg and uniform density is pulled and released. With what frequency does it oscillate?

2. Relevant equations
if i just use f=1/2pi*sqrt(k/m). its the frequency when i hang a weight of 0.1kg at the bottom, but in this case, the weight is distributed along the spring, if we use the idea of center of mass just for checking, the mass will be hung half way of the spring, which means frequency will definitely be different from the former case

3. The attempt at a solution
at equilibrium point, E(total)=E(kinetic)=$$\sum$$(0.5mv^2), (sum of Ekinetic for all points on the spring),since the v=wr where w is the angular velocity(constant for all small segments in the spring), r is radius of moving, which is the amplitude in this case.
so if Ekinetic=$$\sum$$.5m (w^2)r^2, integrate with respect to r,and evaluate form r=0 to r=r.
so Etotal=0.5m(w^2)1/3*r^3,
and at max aplitude, Etotal=Epotential=0.5kx^2=0.5k*r^2
so i got m(w^2)*1/3*r=k.....cant not solve it since r is not given ,and it shouldnt matter..

Last edited: Oct 28, 2007
3. Oct 27, 2007

### Staff: Mentor

Is this homework? Are you supposed to derive the equation?

4. Oct 27, 2007

i am supposed to derive it,can anyone tell my where to start.ran out of ideas...

5. Oct 28, 2007

### rl.bhat

Since the spring constant is given in the problem, the length of the spring is irrelevent.
Use the formula f = (1/2pi)(k/m)^1/2

Last edited: Oct 28, 2007
6. Oct 28, 2007

### eep

What's the force due to a spring look like?

7. Oct 28, 2007

### lilphil1989

response to rl.bhat

i think this should be f=(1/2pi)sqrt(k/m)

as f*2pi=w

and w^2 = k/m

8. Oct 28, 2007

if i just use f=1/2pi*sqrt(k/m). its the frequency when i hang a weight of 0.1kg at the bottom, but in this case, the weight is distributed along the spring, if we use the idea of center of mass just for checking, the mass will be hung half way of the spring, which means frequency will definitely be different from the former case

9. Oct 28, 2007

### rl.bhat

When you consider the mass of the spring f = 1/2pi sqrt[k/(M + m/3)]
where M is the mass attached to the spring and m is the mass of the spring.

10. Oct 28, 2007

### Alana

You have to get dU and dE(kinetic) in terms of density (m\L) and ω (or an equivalent) respectively, then use integration to get U and E(k). Then use E(total) = U + E(k) to solve for the ω, and that's your frequency. You should end up with ω relating to k, L and m.