Solutions manual for mechanics and thermodynamics
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81
4.A particle that hangs from a spring oscillates with an angular frequency !. The spring-particle system is suspended from the ceiling of an elevator car and hangs motionless (relative to the elevator car), as the car descends at a constant speed v. The car then stops suddenly. Derive a formula for the amplitude with which the particle oscillates. (Assume the mass of the spring is negligible.) [Serway, 5th ed., pg. 415, Problem 14]
SOLUTION
The total energy is
E = K + U = 12mv2 + 12kx2
When v = 0, x = A giving
E = 12kA2
which is a constant and is the constant value of the total energy always. For the spring in the elevator we have the speed = v when x = 0. Thus
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Thus
A2 = mk v2
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but ! = mk giving !2 = mk or mk = !12
A2 = v2 !2
v A = !
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CHAPTER 12. OSCILLATIONS |
5.A large block, with a second block sitting on top, is connected to a spring and executes horizontal simple harmonic motion as it slides across a frictionless surface with an angular frequency !. The coe±- cient of static friction between the two blocks is s. Derive a formula for the maximum amplitude of oscillation that the system can have if the upper block is not to slip. (Assume that the mass of the spring is negligible.) [Serway, 5th ed., pg. 418, Problem 54]
SOLUTION
Consider the upper block (of mass m),
F = ma
=sN
=smg
so that the maximum acceleration that the upper block can experience without slipping is
a = sg
the acceleration of the whole system is (with the mass of the lower block being M)
F= (M + m)a
=°kx
The maximum acceleration occurs when x is maximum, i.e. x = amplitude = A, giving the magnitude of a as
kA a = M + m
q
But ! = k giving a = A!2 = sg, i.e.
M+m
A = sg !2
83
6.A simple pendulum consists of a ball of mass M hanging from a uniform string of mass m, with m ø M (m is much smaller than M). If the period of oscillation for the pendulum is T , derive a formula for the speed of a transverse wave in the string when the pendulum hangs at rest. [Serway, 5th ed., pg. 513, Problem 16]
SOLUTION
The period of a pendulum is given by
s
T = 2º L g
where L is the length of the pendulum. The speed of a transverse wave on a string is rø
v =
where ø is the tension and is the mass per unit length. Newton's law gives (neglecting the mass of the string m)
F= Ma
ø° Mg = 0
ø = Mg
and the mass per unit length is |
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but T 2 = 4º2 L or L = |
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v = s |
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CHAPTER 12. OSCILLATIONS |
Chapter 13
WAVES - I
85
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CHAPTER 13. WAVES - I |
Chapter 14
WAVES - II
87
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CHAPTER 14. WAVES - II |
1.A uniform rope of mass m and length L is suspended vertically. Derive a formula for the time it takes a transverse wave pulse to travel the length of the rope.
(Hint: First Ønd an expression for the wave speed at any point a distance x from the lower end by considering the tension in the rope as resulting from the weight of the segment below that point.) [Serway, 5th ed., p. 517, Problem 59]
SOLUTION
Consider a point a distance x from the lower end, assuming the rope
has a uniform linear mass density = mL . The mass below the point is
m = x
and the weight of that mass will produce tension T in the rope above
T = mg = xg
(This agrees with our expectation. The tension at the bottom of the rope (x = 0) is T = 0, and at the top of the rope (x = L) the tension is T = Lg = mg.)
The wave speed is
v = s |
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The speed is deØned as v ¥ dxdt and the time is dt = dxv . Integrate this to get the total time to travel the length of the rope
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2.A uniform cord has a mass m and a length L. The cord passes over a pulley and supports an object of mass M as shown in the Øgure. Derive a formula for the speed of a wave pulse travelling along the cord. [Serway, 5 ed., p. 501]
L - x
x
M
SOLUTION
The tension T in the cord is equal to the weight of the mass M or
X
F = Ma
T° Mg = 0
T = Mg
q
The wave speed is v = T where is the mass per unit length
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v = |
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CHAPTER 14. WAVES - II |
3.A block of mass M, supported by a string, rests on an incline making an angle µ with the horizontal. The string's length is L and its mass is m ø M (i.e. m is negligible compared to M). Derive a formula for the time it takes a transverse wave to travel from one end of the string to the other. [Serway, 5th ed., p. 516, Problem 53]
L
M
θ
SOLUTION |
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The wave speed is given by v = |
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where T is the tension in the |
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length = m |
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string and is the mass per unit q |
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use Newton's laws as shown in the Øgure below.