If we look at circular motion edge on, what we see looks like simple harmonic motion.
An object executing circular motion seems to move back and forth in this view, a cyclic motion very much like simple harmonic motion.
We are discussing the spring force and simple harmonic motion (SHM).
Consider a mass attached to a spring, initially at rest, but pulled
a distance A from the equilibrium position. In this position, the
energy of the system is all potential energy and equals:
Ei = (1/2)kA2.
Now we let go of the spring. At a later time, the energy of the
system is the sum of the potential energy and kinetic energy:
Ef = (1/2)mv2 + (1/2)kx2.
Since the spring force is conservative, and the table is
frictionless, energy is conserved, Ei = Ef.
(1/2)kA2 = (1/2)mv2 + (1/2)kx2.
Solving for v yields:
v = +/- Sqrt{(k/m)(A2 - x2)}.
A mass-spring system oscillates with an amplitude of 3.5cm. If the spring constant is 250 N/m and the mass is 0.50kg, determine (a) the mechanical energy of the system, (b) the maximum speed of the mass, and (c) the maximum acceleration.
(a) The mechanical energy is equal to the potential energy of the
spring at the maximum displacement (since v=0 at this point).
E = (1/2)kA2 = (1/2)(250 N/m)(0.035 m)2 = 0.15
J.
(b) The maximum speed occurs at the equilibrium position (x=0), at
which point all the mechanical energy is now kinetic energy, since
the potential energy is zero. Therefore (1/2)mv2 = E, or,
solving for v:
v = +/- Sqrt{2E/m} = +/- Sqrt{2(0.15J)/(0.50 kg)} = 0.77 m/s.
(c) The maximum acceleration occurs at the maximum displacement
where:
a = (k/m)A = (250 N/m)(0.035 m)/(0.50 kg) = 17.5 m/sē
If we look at circular motion edge on, what we see looks like simple harmonic motion.
An object executing circular motion seems to move back and forth in this view, a cyclic motion very much like simple harmonic motion.
The similarity is deeper than simply that the motion is cyclic. Let's determine the relation between the x coordinate and the x component of velocity, vx, for an object moving in a circle. As shown in the drawing on the right, the radius of the circle will equal the amplitude of the equivalent simple harmonic motion, A. The two triangles shown in the drawing are similar triangles -- the angles are the same and the corresponding sides are scaled by a constant.
The side of the blue triangle corresponding to vx of the red triangle is y = Sqrt{A2 - x2}. Therefore, vx = C Sqrt{A2 - x2}, where C is a constant. But this is the same as the relation derived above, with C replaced by Sqrt{k/m}.
Continuing the similarity between SHM and circular motion, we define three additional quantities:
The tangential speed, v0, equals the maximum speed of the object undergoing SHM.
Conservation of energy gives us the relation (1/2)mv20 = (1/2)kA2, giving us A/v0 = Sqrt{m/k}.
Inserting this expression into the expression for T yields:
T = 2pSqrt{m/k}.
The frequency is the number of cycles per second, or 1 cycle divided by the time required for 1 cycle, the period. Therefore, f = 1/T.
Finally, the angular frequency is defined as w = 2pf = Sqrt{k/m}. The above relation can be seen by again comparing SHM to circular motion. In circular motion, w is the angular velocity, the number of radians traversed per second. Since a complete cycle (one revolution) equals 2p radians, we arrive at the relation between w and f.
Yet again, we go back to the circle.
The x coordinate of the object is related to the angle from the x axis by x = A cosq.
If the angular speed is w, then the angle is given by q = wt.
Therefore
x = A cos(wt)
or by substituting for w,
x = A cos(2pft).
A relation of this type is called sinusoidal. Note that the argument of the cosine function is in radians. When calculating a value, you must set your calculator to radians mode, or convert from radians to degrees yourself.
Given that x = A cos(wt) is a sinusoidal function of time, show that v (velocity) and a (acceleration) are also sinusoidal functions of time.
By inserting the above expression into the relation between velocity and displacement, Eq. 13.6, we get:
v = -wA sin(wt) = -2pfA sin(2pft).
We can likewise substitute into Eq. 13.2 to get the acceleration:
a = -(k/m)x = -w2A cos(wt).
A 2.00 kg mass is attached to a spring with spring constant 5.00 N/m. The mass is displaced 3.00 m in +x, and released from rest. What is the position, velocity and acceleration 3.50 s after the mass is released?
Clearly we will need to make use of the above equations for x, v, and a, and each of them requires the constants w and A. The amplitude is simply equal to the maximum displacement, which will be the initial displacement, since the mass is released from rest, A = 3.00 m. The angular frequency is found from w = Sqrt{k/m} = Sqrt{(5.00 N/m)/(2.00 kg)} = 1.58 rad/s.
Now put these in the appropriate equations to find x, v, and a.
x = A cos(wt) = (3.00 m)cos{(1.58 rad/s)(3.50s)} = 2.20 m
v = -wA sin(wt) = -(1.58 rad/s)(3.00 m)sin{(1.58 rad/s)(3.50 s)} = 3.24 m/s
a = -w
For small angles the motion of a pendulum is also simple harmonic, meaning that the restoring force is proportional to the displacement.
The restoring force is the component of gravity directed perpendicular to the string of the pendulum:
Fres = -mg sinq.
The displacement of the pendulum from equilibrium is the arc length, s = Lq, where L is the length of the pendulum.
As written above, the restoring force is not proportional to s.
However, for small angles sinq = q, resulting in:
Fres = -mgq = -(mg/L) Lq = -(mg/L)s
which is proportional to the displacement.
For the pendulum, the quantity (mg/L) plays the role of the spring constant, k.
The period of a pendulum can be deduced by substituting (mg/L) for k:
T = 2pSqrt{m/(mg/L)} = 2pSqrt{L/g}.
The interesting property of this result is that the period of a pendulum executing small oscillations is independent of the mass, and only depends on its length.
This property was very useful for the making of clocks based on the swings of a pendulum.
What is a wave? A wave is a disturbance in a medium that transports energy, but not matter. For example, you hear someone talking because a sound wave propagates through the air. The air itself doesn't move in bulk -- that would cause a breeze -- but oscillations of pressure are transported through the air. Waves of water don't transport the water, that is, they don't produce a current. The waves on a string move the string transversly, but do not produce a net displacement of the whole string.
Waves are transported by a medium such as air, water, or a string. (A special exception is electromagnetic waves which propagate without the presence of a medium.) There are two principle types of waves, distinguished by how elements of the transporting medium move:
We represent the disturbance of the wave by graphs of the displacement of the elements of the medium at a particular time. The "fundamental" waves that we will work with have graphs that are sinusoidal in shape. The maximum positive displacements are called crests or peaks and the maximum negative displacements are called troughs.
Although waves can be carried in a variety of mediums, they are characterized by the same quantities, frequency, amplitude, and wavelength. To picture "amplitude", imagine a sinusoidal wave traveling down a string under tension. Pick a small piece of the string, maybe make it a different color from the rest, and watch its motion. That piece of string will move up and down in an oscillatory motion as the wave passes, in fact the motion is simple harmonic. The piece of string has an equilibrium position, its location when the string is not vibrating, and in its simple harmonic motion, the piece of string has a maximum displacement from the equilibrium position. This maximum displacement is the amplitude of the wave, A.
The period of the wave, T, is the time it takes for this piece of string to make a complete cycle. And as was true for SHM, the frequency is the inverse of the period, f = 1/T.
The wavelength, l (Greek letter lambda), is the distance along the string between successive peaks of the wave. It takes a time T for a peak to move by one wavelength (diagram). Therefore, the velocity of the wave is v = l/T = fl.
The speed of a wave traveling along a string is determined by the tension, F, in the string and the mass per unit length, m, of the string, v = Sqrt{F/m}. Dimensional analysis shows that the units of the right hand side are length/time, the same as velocity.
A wave traveling in the positive x direction has a frequency of 25.0 Hz, as in Figure P13.32. Find the (a) amplitude, (b) wavelength, (c) period, and (d) speed of the wave.
(a) The amplitude is half the peak to peak distance of 18cm, or A = 9.0cm.
(b) The 10 cm distance shown in the figure is one half a complete cycle, from a crest to a trough, thus wavelength is twice this distance, or l = 20 cm
(c) Since f=25.0 Hz, T = 1/f = (1/25)s = 0.0400 s.
(d) Use v = fl = (25.0 Hz)(0.20 m) = 5.0 m/s.