hw 10 - fa14 - oscillations and waves

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1 1 1 Equation Chapter 1 Section 1 HW 10 - fa14 - oscillations and waves : Chapter 15 : 2, 14, 26, 32, 33, 42, 51. Chapter 16 : 8, 21, 22. Pool of problems to study : 15.2, 15.26, 15.33, 15.42, 16.21. chapter 15, problem 2 : A body undergoes simple harmonic motion of amplitude and period . (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant? (a) maximizing both sides of newton’s 2 nd law, 212\* MERGEFORMAT (.) Plugging numbers into 12, 313\* MERGEFORMAT (.) (b) We take a moment to derive a very fundamental relation about the spring constant. Fact : If is the position-function for oscillatory motion 1 of angular frequency , then Using Newton’s 2 nd Law upon a particle of mass subject to a spring-force , in which , we have, 414\* MERGEFORMAT (.) Specializing 14 to our physical situation, we calculate the spring-constant as, 1 The most general possible expression of oscillatory motion of angular frequency is ; it is upon this expression that the operator acts to produce . This is closely-related to the identity ; for , we notice that procures the property acting to produce . This bit of mathematical trickery will definitely be encountered again in your engineering courses (i.e., the property is a major calculational-convenience).

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HW 10 - fa14 - oscillations and waves: Chapter 15: 2, 14, 26, 32, 33, 42, 51. Chapter 16: 8, 21, 22. Pool of problems to study: 15.2, 15.26, 15.33, 15.42, 16.21.

chapter 15, problem 2: A body undergoes simple harmonic motion of amplitude and period. (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant?

(a) maximizing both sides of newtons 2nd law,

Plugging numbers into ,

(b) We take a moment to derive a very fundamental relation about the spring constant. Fact: If is the position-function for oscillatory motion[footnoteRef:1] of angular frequency , then Using Newtons 2nd Law upon a particle of mass subject to a spring-force , in which , we have, [1: The most general possible expression of oscillatory motion of angular frequency is ; it is upon this expression that the operator acts to produce . This is closely-related to the identity ; for , we notice that procures the property acting to produce . This bit of mathematical trickery will definitely be encountered again in your engineering courses (i.e., the property is a major calculational-convenience).]

Specializing to our physical situation, we calculate the spring-constant as,

We have also showed that is also an acceptable way to calculate the spring-constant, due to the insight afforded by the machinations and .

Chapter 15, problem 14: A simple harmonic oscillator consists of a block of mass attached to a spring of spring constant. When, the position and velocity of the block are and. Be careful to note our notation: in a spring, and , and is the displacement at and is the velocity at .

(a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at? (a) A very handy placeholder variable to calculate is the total energy in the spring (which is independent of time), which appears as,

Hence, the amplitude of the oscillations (maximum displacement) must be given by the displacement when the spring has its potential energy take on its maximum value , when ,

(b) The position at is yielded by the explicit ; hence, we must use (a 1st placeholder-variable) and (a 2nd placeholder variable), and we must also[footnoteRef:2] calculate[footnoteRef:3] , [2: We have calculated a number of intermediary numerical values that is greater than I like to do, but since oscillation-problems involve trigonometric (transcendental) functions, one can easily make a giant symbolic mess by being too bent on symbolic answers. Therefore, we can calculate intermediaries.] [3: We used the property , which means . Notice that this means the equation has two possible values for the phase , the other being ; hence this problem is, in fact, ill-posed, and does not have a unique answer. Probabilistically, at least half (65) of my 130 students in PHYS 2048 should have emailed me questions about this problem.]

We pause for a moment, and notice that we have all quantities known in , so,

Afterword: We noted in the footnote that the solution to this problem is non-unique. Therefore, using the other possible value for the phase-angle , we have,

chapter 15, problem 26: In the Figure, two blocks (and) and a spring () are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is. What amplitude of simple harmonic motion of the spring/blocks system puts the smaller block on the verge of slipping over the larger block?

The condition of interest is , in which , in which is the coordinate of the combined system , and hence , in which . We also have , in which . Let the acceleration be at its maximum value, at which the amplitude would be on the verge of being sufficiently large to cause any slippage. Then, , and,

Chapter 15, problem 32: The Figure shows the kinetic energy of a simple harmonic oscillator versus its position . The vertical axis scale is set by. What is the spring constant?

The spring constant is given by the total energy of the spring,

Chapter 15, problem 33: A block of mass, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant. A bullet of mass and velocity of magnitude strikes and is embedded in the block (see figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

The amplitude of the resulting simple harmonic motion is given by,

Plugging numbers into , we have,

Chapter 15, problem 42: Suppose that a simple pendulum consists of a small bob at the end of a cord of negligible mass. If the angle between the cord and the vertical is given by where and what are (a) the pendulums length and (b) its maximum kinetic energy?

For a pendulum consisting of mass , Newtons 2nd Law in the y-direction, which is gotten via (implied by the equation on any circle of radius ) yields,

Hence, the pendulum length is easily found to be . The maximum kinetic energy occurs when the mass has maximum angular speed,

Chapter 15, problem 51: In the Figure, a stick of length oscillates as a physical pendulum. (a) What value of distance x between the sticks center of mass and its pivot point O gives the least period? (b) What is that least period?

The period of a physical pendulum of inertia , total mass in a gravitational field about a rotation-axis a distance away from the objects center of mass is[footnoteRef:4] . Proof: consider an arbitrary body with these indicated dimensions, [4: Proof:]

Newtons 2nd law in the radial direction yields,

For our problem, the parallel axis theorem yields a moment of inertia of . We are to extremize the period calculated from , ,

This expression looks complicated, but since it is equal to 0, we can use a trick. Consider the equation ; as long as , the solution to the equation is ; in this case, is solved by (see the numerator), implying . In turn, the value of the period at is given by,

Afterword: Although not part of the Pool of Questions, this is a somewhat-popular exam-question.

Chapter 16, problem 8: The Figure shows the transverse velocity versus time t of the point on a string at, as a wave passes through it. The scale on the vertical axis is set by. The wave has the form. What is ? (Caution: A calculator does not always give the proper inverse trig function, so check your answer by substituting it and an assumed value of into and then plotting the function.)

The disclaimer in the problem-statement should have also been in Chapter 15, problem 14. That is, the authors notice non-uniqueness here, but not when it is more subtle in Ch 15, pr 14. The fact of the matter is that people make mistakes, and you need to develop your own solutions to physics problems. Another example of a gross error on the solution manuals part is Chapter 7, problem 42. Anyway: we write,

Using the condition , and the information , we have,

Chapter 16, problem 21: A wire is held under a tension of with one end at and the other at. At time, pulse 1 is sent along the wire from the end at . At time, pulse 2 is sent along the wire from the end at. At what position x do the pulses begin to meet? Preliminary: We require the velocity equation: the velocity of any wave (pulse, oscillation, etc.) propagating through a string of mass-density (here ), under tension . A transverse wave is a pulse along a string. Let the amplitude of the pulse be , and consider the circle approximately traced out by this radius .

In the limit of , the angles become , but the tensions stay the same. Taking the component lying parallel to , the sum of the forces is,

This force-sum elicits a centripetal acceleration,

In this problem, each pulse has the same speed , but the 1st pulse has an x-coordinate, , given by kinematics as at time , where the 2nd pulse has an x-coordinate, given by . . Pretending the pulses are point-objects (i.e., of zero dimensionality), and letting them meet at time (i.e., is a placeholder-variable), we have,

Calculating the position at this time, we have,

Chapter 16, Problem 22: A sinusoidal wave is traveling on a string with speed. The displacement of the particles of the string at varies with time according to , where , , and . The linear density of the string is. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form , what are (c) ym, (d) k, (e) , and (f ) the correct choice of sign in front of v? (g) What is the tension in the string? Solution: (a) . (b) . (c) (as indicated). (d) . (e) (as indicated). (f) The phase velocity at is evidently given by ; because , we realize , and since the phase velocity is , in which , we have . Hence, the phase velocity is positive. The function representing a positively-propagating wave is . (g) solve,