Test 2 trial

course Phy 202

July 17 12am

Problem Number 1 If the peaks of a traveling wave are separated by 4 meters, and if the wave propagates at 186 m/s, what is the frequency of the wave? What is the period of a cycle of the wave?

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.f= v/wvl

F= (186 m/s)/4 m

F= 46.5 Hz

P= l/v

P= (4m)/(186 m/s)

P= 0.022 sec

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Problem Number 2

A string defines the x axis, with the origin at the left end of the string. The string has tension 7 Newtons and mass per unit length is 3 grams / meter. The left-hand end of the string is displaced perpendicular to the string in a cyclical manner in order to create a traveling wave in the string. At clock time t the position of the left-hand end of a long string is y = 1.13 cm * sin ( ( 8 `pi rad/s) t ). Explain where the energy is in this wave, and find the energy per unit length of the wave.

.For this problem I am a bit lost. With the information I am given, I am able to find the velocity, though I’m not sure that’s what I need.

V= `sqrt(7n)/(3g/m)

V= 1.53 m/s

I’ve searched through my notes and my book and can’t seem to find how to find the energy per unit length of a wave..

This is covered in the assigned Introductory Problem Sets, Problem 17. Here's a copy of the current version:

Set 56 Problem number 17


Problem

The mass density of a string is 16.7 grams/meter, its length is 35 meters and its tension

is such that the velocity of a transverse wave in the string is 54 m/s.  How much

energy is there in the string if it carries a traveling wave with amplitude 1.95 meters and

frequency 128 Hz? 

College and University Physics students also answer the following:

Solution

Each particle in the string undergoes SHM with

amplitude 1.95 meters and frequency 128 Hz.  From the mathematic of SHM, or simply from

the circular model, the maximum velocity of each particle is therefore

You will recall that the total energy of any mass

in SHM is constant throughout the motion, with all the energy potential at the extreme

points of the motion and all the energy kinetic at the equilibrium point, where the

velocity is maximum.  Since every particle is in SHM with the same frequency and

amplitude, though different particles are at different points and have differing mixes of

KE and PE, the total energy of every particle is equal to the KE it has at its maximum

velocity.  We can therefore state that the total energy of the wave motion in the

string is the energy its total mass would have at the maximum velocity:

where vMax is the maximum SHM velocity of a

particle.

The total mass of the string is simply 16.7

grams/meter * 35 meters * (1 kg / (1000 grams) ) = .5845 kg. We have already found vMax.

  Thus the total energy of the string is

The power required to maintain the wave is the

energy supplied per unit of time.  Each time the wave travels the length of the

string, the 717616.7 Joules in the string will pass on beyond the end of the string (perhaps to

another string, perhaps into some mechanism that makes use of the energy, perhaps

dissipated into some medium--it doesn't matter where it goes, it just goes). 

University Physics students note:  To be

completely rigorous we should consider a Riemann sum of the energies over small

increments.  A typical length increment `dx of the string will have its SHM defined

by a sample point xi in the ith interval.  Its total energy will be equal to that of

the mass 16.7 g/m * `dx in the interval, and will hence be .5( 16.7 g/m * `dx ) * ( 1567 m/s)^2 =

4.100667E+07 kg / s^2 * `dx.  When this quantity is summed over all intervals the 4.100667E+07 kg/s^2

factors out, the remaining increments `dx add up to the length of the string and we obtain

4.100667E+07 kg/s^2 * 35 meters = 717616.7 Joules.  This result does not depend on the increments

`dx, but the approximation that all the mass in `dx moves with the same SHM is not

completely valid.  However, as the interval length `dx approaches zero, the

approximation becomes more and more valid, and becomes exact in the limit.

Generalized Solution

A mass m moving in SHM with amplitude A and

frequency f has angular frequency `omega = 2 `pi f, from which we see that its maximum

velocity is `omega A = 2 `pi f A.  Its kinetic energy at this maximum velocity is

therefore

This is the same as the total energy of the mass at

any point.  Thus every mass increment of the string has total energy given by this

expression, and when the resulting energies are totaled for the entire string we obtain

The total mass of the string is `mu * L, where L is

string length and `mu is mass per unit length.  Thus

If the wave moves with velocity v, the total energy

of the wave will pass beyond the string in time

so we must supply energy at the rate

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Problem Number 3

By how many decibels does a sound of intensity 7 * 10^ 10 watts/m^2 exceed a sound of intensity 4 * 10^ 4 watts/m^2?

.dB= 10 log (I/Io)

dB= 10 log (7*10^10)/(10^-12)

dB= 10 log (7*10^22)

dB= 228.45 watts/m^2

dB=10 log (4*10^4)/(10^-12)

dB= 10 log (4*10^16)

dB= 166.02 watts/m^2

228.45-166.02= 62.43 So the sound of intensity 7*10^10 watts/m^2 is 62.43 dB stronger than a sound of intensity 4*10^4 watts/m^2

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Problem Number 4

What is the fundamental frequency of a longitudinal standing wave in an aluminum rod of length 7 meters, balanced at its midpoint, if a longitudinal disturbance travels at 5000 m/s?

.frequency= velocity/ 2L

Since the wave is balanced at its midpoint it has a node at its midpoint.

Since its ends are free it has antinodes at the ends.

The distance between a node and an adjacent antinode is 1/4 wavelength.

Thus the length of the rod is L = 2 * (1/4 wavelength) = 1/2 wavelength, and the wavelength is therefore double the length of the rod.

frequency = velocity / wavelength, and in this case the wavelength is 2 L. So your equation .frequency= velocity/ 2L is correct for this situation. However it's not a general equation, but results from the general relation frequency = velocity / wavelength in a way you need to document. In other words, you do need to explain why the wavelength is 2 L.

This question could easily have been phrased with the rod balanced at two different points, for example two points each 1/4 of the way from an end. So on many versions of this problem, you're going to need to reason out the wavelength.

Frequency= (5000 m/s)/ 2*7

Frequency= 357.14 s

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Problem Number 5

A string of length 8 meters is fixed at both ends. It oscillates in its second harmonic with a frequency of 156 Hz and amplitude .3 cm. What is the equation of motion of the point on the string which lies at 1.1 meters from the left end? What is the maximum velocity of this point?

.Im having trouble finding an equation of motion that works for this problem. The equation I found was x(t)= Acos(2’pi*f*t+ ‘theta)

I’m so lost! I know A= amplitude, f= frequency, but I don’t know what t is or ‘theta. Is this the right equation to use?

The reasoning you need in order to do this problem is also covered in the Introductory Problem Sets.

Briefly, the second harmonic has configuration N A N A N, consisting of four quarter-wavelengths, distributed along its 8-meter length. So the wavelength is 8 meters.

A graph of the string with each point at its maximum displacement from the equilibrium position is

y = A sin( k x ), where A is the amplitude of the oscillation and k x changes by a full cycle of 2 pi when x changes by a wavelength. Thus k * 8 m = 2 pi rad and k = 2 pi rad / (8 m) = (pi / 4 rad / m).

The amplitude of motion as position x is therefore A sin( k x ) = .3 m sin( pi/4 rad/m * x).

Each point undergoes SHM with frequency 156 Hz, so that its angular frequency is 156 cycles/sec * 2 pi rad/cycle = 312 pi rad/sec.

The point at position x therefore undergoes SHM of amplitude .3 m sin(pi/4 rad/m * x), and frequency 312 pi rad/sec. Its equation of motion is

y(x, t) = .3 m sin(pi/4 rad/m * x) * sin(312 pi rad/sec).

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Problem Number 6

A candle flame 1.67 cm high is located 11 cm in front of a circular mirror. A real inverted image forms 3 cm from the mirror.

• What is the focal length of the mirror?

• How large is the image?

• Sketch a ray diagram explaining how the image is formed.

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Focal length

1/f= 1/d0+ 1/di

1/f= (1/11 cm)+ (1/3 cm)

1/f= 0.424 cm

F= 1/0.424 cm

F= 2.36 cm

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.Image size= ho/hi

1.67 cm/ 3 cm= 0.56 cm

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Good work on most questions.

See my notes and let me know if you have additional questions.