Elasticity and Oscillations. Intrinsic Strength of Materials. Hanging by a hair Can a single hair support your weight? Steel wire? A hair, cable etc. can hold more weight if it has greater cross section. Intrinsic Strength of Materials.
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your weight? Steel wire?
A hair, cable etc. can hold more weight if it has greater cross section.
Steel cable is intrinsically stronger than hair, rope
Stress = F/A
Apply a force to both ends of a long wire. These forces will stretch the wire from length L to L+L.
F = kΔL
If we pull on a spring it increases in length
If we pull on a wire increases in length.
Longer wires (and springs) will stretch more than
The fractional change in length
Force per unit cross-sectional area
Hooke’s Law (Fx) can be written in terms of stress and strain (stress strain).
The spring constant k is now (F=kx)
Y is called Young’s modulus and is a measure of an object’s stiffness. Hooke’s Law holds for an object to a point called the proportional limit.
Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional area 1.0106 m2, has a Young’s modulus of 2.0109 Pa. By how much must you stretch a guitar string to obtain a tension of 20 N?
If the stress on an object exceeds the elastic limit, then the object will not return to its original length.
An object will fracture if the stress exceeds the breaking point. The ratio of maximum load to the original cross-sectional area is called tensile strength.
The ultimate strengthof a material is the maximum stress that it can withstand before breaking.
Example (text problem 10.10): An acrobat of mass 55 kg is going to hang by her teeth from a steel wire and she does not want the wire to stretch beyond its elastic limit. The elastic limit for the wire is 2.5108 Pa. What is the minimum diameter the wire should have to support her?
f=1/T T = 1/f
Frequency is measured in Hertz = 1 cycle /second
We can plot the position of the ball as a function of time. A is the distance from the bottom of the cup.
Simple harmonic motion (SHM) occurs when the restoring force (the force directed toward a stable equilibrium point) is proportional to the displacement from equilibrium.
The motion of a mass on a spring is an example of SHM.
The restoring force is F = kx.
When the stretch is a maximum, a will be a maximum too.
The velocity at the end points will be zero, and it is a maximum at the equilibrium point.
When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system.
As soon as m released v becomes negative.
t=1/4 T: equilibrium x=0, v max(-), a=0
t= ½ T: x=-A, v=0 a positive, begins + motion
t= ¾ T: x=0 equilibrium v max (+), a=0
t= T; back at the beginning. Here we go again.
At equilibrium x=0 v max
v2max =k/m A2 v= √k/m A
a =-ω2 x
where A is the amplitude of the motion, the maximum displacement from equilibrium, A = vmax, and A2 = amax.
where is the angular frequency of the oscillations, k is the spring constant and m is the mass of the block.
Example (text problem 10.30): The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the equilibrium point?
At equilibrium x = 0:
Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = vmax = A.
The amplitude A is given, but is not.
A mass on a spring has amplitude A. If A is doubled, the total energy of the system is:
C) the same.
E) 1/4 as much.
Example (text problem 10.41): The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of 2.0 kHz by moving back and forth with an amplitude of 1.8104 m at that frequency.
(a) What is the maximum force acting on the diaphragm?
The value is Fmax=1400 N.
Since mechanical energy is conserved, E = Kmax = Umax.
The value of k is unknown so use Kmax.
The value is Kmax= 0.13 J.
Example (text problem 10.47): The displacement of an object in SHM is given by:
What is the frequency of the oscillations?
Comparing to y(t) = A sint gives A = 8.00 cm and = 1.57 rads/sec. The frequency is:
Other quantities can also be determined:
The period of the motion is