Simple Harmonic Motion Questions



Solving Oscillation Questions

1. If there is a spring involved, find the spring constant k

There is probably also a mass given.

a) If it is at rest, at equilibrium, use forces ((F = 0)

b) If it is in motion, use conservation of energy

2. If you need time, find the frequency and then the period using the formula.

3. If you need velocity, use conservation of energy.

Conservation of energy is really easy if you have the amplitude A (maximum displacement).

4. If you need to find distance, use the Amplitude and your common sense.

5. Make sure your calculator is in RADIANS when you use 2(f in trig functions.

6. Pendulums are similar.

Simple Harmonic Motion Questions

(Giancoli p 303-304 # 2, 4, 6 ,7, 8, 11,12,14,18, 27 – 3rd ed.)

Giancoli p 327-328 # 2, 4, 6 , 8, 9, ,12,15,22, 31 – 4th ed.

#12. – illustrative of some good principles: (numbers have changed in 4th edition text!)

A mass of 1.80 kg stretches a vertical spring by 0.275 m. If the spring is stretched an additional 0.110 m and released, how long does it take to reach the (new) equilibrium position?

m = 1.80 kg x = 0.275 m

What can we find out with these two quantities? (dead silence!)

What do you think of when you think of a spring? K !!! A spring constant. How stiff it is.

Using Forces

What is the first adjective? Vertical!

Fg = Fe

mg = kx

k = mg/x

( Where did the negative sign go?

Using Energy

initial = final

Eg = Ee

mgh = ½ kx2

mgx = ½ kx2

k = 2mg/x

( Why are the two answers different by a factor of two?

Let’s answer both together.

Looking at Forces:

1. With SHM, the acceleration is always changing. So how can you use (F = ma ?

Perhaps at some instant, we could determine a somehow, and then say that (F = Fe + Fg.

Note that (F is not constant; not only is it continually changing, but it also switches signs!

We can use F=ma when we have the oscillator stabilized at the equilibrium position. The (massless) spring is some length; when a mass is attached, it stretches by a distance x0 . This is the new equilibrium position.

Drawing a free-body diagram for the mass (not the spring) in this position:

The upwards elastic force (+Fe) is equal and opposite to the force of gravity (–Fg).

Since a = 0, (F = 0. Fe – Fg = 0

( kxo – mg = 0 (k = mg/x

This answer for k is correct.( (This is the best way to find k, unless you have f )

(Remember this!)

Looking at Energy:

It is true that the energy stored in the spring is ½ kx2 . Is the initial energy mgh? Well this only works if we release the mass from the rest point (x = 0). The mass will not stop at the new equilibrium position (x = x0), but will keep on going twice as far, then stop and come back up, and keep oscillating. It will never be at rest at x = x0 … unless energy is dissipated and the oscillations are damped down. So the equation Eg = Ee is not true at x = 0.275 m (the rest position). You have to also include the kinetic energy.

Ok. We have k , so now what do we do?

We have the formula for frequency – which interestingly does not depend on displacement.

Find f (from m and k)

What exactly does f tell us? Cycles per second. Ok, find T. (# of seconds for 1 cycle).

We want the time for ¼ period. This is ¼ of T.

Voila!

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download