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A mass on a spring will oscillate if the mass is pushed or pulled from its equilibrium position. Why?
We saw from the Hooke’s Law experiment that the force of a spring is related to how far the spring is pulled or pushed from equilibrium: Fspring = -kx where the minus simply indicates that if you push in on the spring, the spring pushes out.
The spring constant, k, describes how “stiff” the spring is. A large k indicates that a large force is needed to stretch the spring.
But why does the mass oscillate?
From Newton’s Second Law, if we ignore other forces like friction or gravity:
S F = ma and F = -kx leads to -kx = ma .
-kx = ma
This says that if x is positive (spring stretched), the force is negative and hence the acceleration is negative. This will produce a velocity that, if initially zero, will become negative, tending to reduce the positive x. As x approaches zero, the equilibrium position, the force approaches zero and the acceleration will approach zero; but this still leaves a negative velocity!
Now as it passes the equilibrium position, the mass has negative velocity so it will pass through the equilibrium position and end up with a negative (compressed) position. Here the force (F=-kx) becomes positive giving a positive acceleration. This will make the negative velocity less negative, but still negative. Hence the mass will go to an even more negative position, with an even bigger negative force and acceleration that will continue to slow it down until it reaches zero speed.
At this point where the speed is zero, we have a negative (compressed) position which gives a positive force and hence positive acceleration. This acceleration will then cause the speed to increase to a positive value which will cause the mass to move back towards equilibrium.
This process continues and we get an oscillation!
Using the calculus, we can solve -kx = ma
(a differential equation: -kx = md2x/dt2) to get: x = A sin(wt + qo) ,
where w =(k/m) comes from substituting the solution back into the differential equation.
This should appear reasonable: the sine function oscillates (goes up and down in value) just like the mass oscillates! But the sine function needs an angle to operate on! Where is the angle in the problem?
x = A sin(wt + qo) .
The sine function is operating on the quantity: (wt + qo). This expression must be an angle. But what angle?
There is no “geometric” angle in the problem because the problem is only in one dimension.
Instead, we call this kind of angle a phaseangle. A phase angle simply describes where in the oscillation the wave is!
The crest of the
sine wave is
located at 90o,
the trough at 270o
and it crosses zero
at 0o, 180o and
at 360o (or 2p radians).
x = A sin() where is a phase angle
in an oscillation; changes withtime(goes2radians in time T)so
q = wt + qo = (2/T)*t + qo
where2/T = = 2pf; and
qois simply the phase angle when t=0 (where the oscillation starts at t=0).
x = A sin(wt + qo) .
The amplitude, A, describes how far up and how far down x goes. Since sine has a maximum value of 1 and a minimum value of -1, A is used to put in units and to give the amplitude of the oscillation.
x = A sin(wt + qo) .
We have seen that w describes how fast the mass oscillates. But what does this oscillation (w) depend on?
By substituting our solution into the differential equation, we get: w = (k/m) .For stiffer springs and lighter masses, the frequency of the oscillation increases.
Note: the Amplitude does NOT affect the frequency!
w = (k/m)
Why doesn’t the Amplitude affect the frequency? The more you stretch the spring (bigger Amplitude), the farther the oscillation has to go. However, you also have bigger forces which mean bigger accelerations and bigger speeds. Which wins, the bigger distance or the faster speeds? The math tells us that it is a tie! The Amplitude does NOT matter here!
For the energy of masses on springs, we have:
Etotal = KE + PE = (1/2)mv2 + (1/2)kx2.
We also know that x = A sin(wt + qo) .
If we look at how x changes with time, we get the speed: v= dx/dt = d[A sin(wt+qo)]/dt =
= wA cos(wt + qo). Therefore, Etotal =
(1/2)mw2A2cos2(wt+qo) + (1/2)kA2sin2 (wt+qo).
Note: when sine is maximum, cosine is zero; and when sine is zero, cosine is maximum!
Does Etotal change with time, or is it a constant?
(1/2)mw2A2cos2(wt+qo) + (1/2)kA2sin2 (wt+qo).
Recall that w = (k/m) . Thus,
(1/2)mw2A2 = (1/2)kA2 and since sin2q+cos2q=1, we see that Etotal = (1/2)mw2A2 = constant .
Note that the energy of oscillation depends on the frequency squared and the amplitude squared.
In the Oscillations Lab in Physics I, we experimentally developed equations to describe the oscillations of a pendulum.
In this case, we have a real geometric angle AND we have a phase angle. The geometric angle (the angle the pendulum makes with the vertical) oscillates, so we might guess a solution:
qpendulum(t) = qmax sin(wt+fo).
In the lab we found that the period depended on the square root of the length, but not on the mass and only slightly on the angle.
If we start from Newton’s Second Law for rotations: St = Ia, we get:
-mg sin(q) L = mL2 d2q/dt2 .
For small angles, sin(q) q (where the angle is measured in radians), so we now have:
- g q = L d2q/dt2 .
- g q = L d2q/dt2But this looks just like
-kx = md2x/dt2 and so has a similar solution: q(t) = A sin(wt + fo) where
w = [g/L] = 2p/T , where the expression for w came from substituting our solution back into the differential equation.
This says that if we don’t have too big an oscillation, the period does not depend on the mass or angle, but only on the L !
If you wiggle one end of a string (or slinky) that has a tension on it, you will send a wave down the string. Why?
We can consider this by looking at Newton’s Second Law:
If we consider the first part of the wave, we see that the Tension on the right has a zero y component, but the Tension on the left is pulling up. Thus, S Fy = Tleft-y + Tright-y > 0.
This will cause this part of the string to start moving upwards! [Note: Tleft-x = Tright-x]
A little later, this part of the string is moving upwards, but now the Tension on the left has less of a y component and the Tension on the right: Tleft-y + Tright-y < 0 . This will tend to slow this part of the string down! [Again, Tleft-x = Tright-x ]
At the top of the pulse, the wave has slowed to zero, but both the left and right Tensions are pulling down, so there will continue to be a negative y force, and this will cause this part of the string to start moving down.
By continuing to look at each part of the string, we can understand how the wave can move down the string.
Starting with Newton’s Second Law for a piece of mass on the string, we have:
SFy = +Tsin(qL) - Tsin(qR) = Dmay .
[here T is the tension in the string.]
Note: for small angles, sin(q) tan(q) q,
so : sin(qL) - sin(qR) = Dsin(q) Dtan(q).
T Dtan(q) = Dm ay .
Further, tan(q) = dy/dx. Next, let’s divide both sides by Dx, so with Dm/Dx = m we have: T d2y/dx2 = m ay , or
T 2y/x2 = m 2y/t2 .
The above equation is called the wave equation for a string. Recall that it comes directly from Newton’s Second Law.
T 2y/x2 = m 2y/t2
The solution to this 2nd order partial differential equation is f(kx wt + qo), but we’ll choose the sine function for our function, f. Fourier Series theory says any periodic function can be expressed as a series of different sine waves, so our choice of the sine function really doesn’t limit us.
y(x,t) = A sin(kx wt + qo) .
Wave Equation: T 2y/x2 = m 2y/t2
Solution: y(x,t) = A sin(kx wt + qo) .
Note that this k is the wave vector, NOT the spring constant!
Note: we also have two T’s: the T for tension and the T for period!
Tension2y/x2 = m 2y/t2
y(x,t) = A sin(kx wt + qo)
If we substitute this solution into our equation, we get: Tensionk2 = mw2 , or
w/k = [Tension/m] .
Does w/k have any special meaning?
q(x,t) = (2/)*x ± (2/Tp)*t= kx ±wt + qo
If we consider how the crest moves (or any other particular phase), then q = constant (= 90o for the crest) and we have for the position of that phase angle in time:
xq = (qwt - qo)/k and so the phase speed = vq = dxq/dt = /k . Also,
vq =distance/time = /Tp = f =/k .
We saw previously that w/k = [Tension /m] , so now we have an expression for the phase velocity of the wave on a string:
vq = lf = w/k = (Ttension)/m)
where m = m/L (mass/length of string).
Note that there are two velocities in this case: the velocity of a particular phase (such as a crest, vq=dxq/dt), and the velocity of a particular part of the string (going up and down, vy=dy/dt).
The velocity in the relations vq =lf and
vq =(Ttension)/m) is the phase velocity (vq=dxq/dt).
At any particular location (x), the mass at that point has a y motion of y = A sin(kx-wt+qo) = -A sin(wt+fo) where the fo=kx+qo=constant and so has a vy = -A cos(wt+fo) .
NOTE: vq and vy are different things!
Note that the phase velocity is related to the wavelength (l) and frequency (f), but it does not depend on these:
vq = lf = w/k .
If you double the wavelength, the frequency will halve, but the phase velocity remains the same.
The phase speed depends on the material it is going through (Tension and m):
vq = [Tension /m].
y(x,t) = A sin(kx t + qo)
sawtooth waves, etc.) can be formed by a
superposition of sine waves - this is called
Fourier Series. This means that sine
waves can be considered as fundamental.
Each part of a nice sine wave oscillates just like a mass on a spring. Hence it’s energy is that of an oscillating mass on a spring. We saw before that this depended on w2A2.
The power delivered by a wave is just this energy per time. The rate of delivery depends on the phase speed of the wave. Thus the power in the wave should depend on: P w2A2v where means proportional to.
Do waves “bounce” off of obstacles? In other words, do waves reflect?
We’ll demonstrate this in class with a slinky.
There are two cases: when waves bounce off a stiff obstacle, the waves do reflect, but they change phase by 180o; when waves bounce off a loose obstacle, the waves do reflect, but they do NOT change phase.
When a wave on a string encounters a fixed end, the reflected wavemust interfere with the incoming waveso as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.
Below is the same situation, only this time we picture the situation at a time (1/4)T later.
When a wave on a string encounters a free end, the reflected wavedoes NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.
Again, we picture the situation at a time
(1/4) T later.
To create what are called standing waves (these should be demonstrated in class), we need the following condition:
#(l/2) = L ;
this says: we need an integer number of half wavelengths to “fit” on the Length of the string for standing waves.
We can vary the wavelength by either varying the frequency or the speed of the wave
(recall: vq = lf where vo = [Tension/m] ) .