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Chapter 16: Problem 56

A standing wave pattern on a string is described by $$ y(x, t)=0.040(\sin 5 \pi x)(\cos 40 \pi t) $$ where \(x\) and \(y\) are in meters and \(t\) is in seconds. For \(x \geq 0,\) what is the location of the node with the (a) smallest, (b) second smallest, and (c) third smallest value of \(x ?\) (d) What is the period of the oscillatory motion of any (nonnode) point? What are the (e) speed and (f) amplitude of the two traveling waves that interfere to produce this wave? For \(t \geq 0,\) what are the (g) first, (h) second, and (i) third time that all points on the string have zero transverse velocity?

Short Answer

Expert verified
Nodes at 0, 0.2, and 0.4 m; period 0.05 s; wave speed is 8 m/s; amplitude 0.02 m. Zero velocity at 0.0125, 0.0375, and 0.0625 s.

Step by step solution

01

Identifying Node Locations

For the node locations, we need to find when the sine function is zero. Nodes occur at positions where \( \sin(5\pi x) = 0 \). Since \( \sin \theta = 0 \) at \( \theta = n\pi \), we set \(5\pi x = n\pi\) and solve for \(x\): \[5\pi x = n\pi \implies x = \frac{n}{5}\]Nodes occur at \( x = 0, \frac{1}{5}, \frac{2}{5}, \ldots \), with increasing \(n\).
02

Smallest Node Locations: A, B, C

Based on the identified formula for node locations:- (a) The smallest node location is at \(x = \frac{0}{5} = 0\).- (b) The second smallest node location is at \(x = \frac{1}{5} = 0.2\) meters.- (c) The third smallest node location is at \(x = \frac{2}{5} = 0.4\) meters.
03

Period of Oscillatory Motion at Non-node Points

The period \(T\) of the wave is determined by \(\cos(40\pi t)\). The relation for the period is \(T = \frac{2\pi}{\omega}\) where \(\omega = 40\pi\). Thus, \(T = \frac{2\pi}{40\pi} = \frac{1}{20}\) seconds.
04

Speed and Amplitude of Traveling Waves

The wave number \(k = 5\pi\) and angular frequency \(\omega = 40\pi\). The speed \(v\) of the traveling waves is \(v = \frac{\omega}{k} = \frac{40\pi}{5\pi} = 8\) meters/second. The amplitude of each traveling wave is half the amplitude of the standing wave, \(0.040/2\) meters, so the amplitude is \(0.020\) meters.
05

Times When All Points Have Zero Transverse Velocity

A point on the string has zero transverse velocity when \(\cos(40\pi t) = 0\). Solving \(40\pi t = \frac{\pi}{2} + n \pi\) gives:1. (g) First time: \(t = \frac{1}{80}\) seconds.2. (h) Second time: \(t = \frac{3}{80}\) seconds.3. (i) Third time: \(t = \frac{5}{80} = \frac{1}{16}\) seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Node Locations
In a standing wave, nodes are points where there is no movement; the amplitude is consistently zero. To find these locations in our given wave function, we need to solve the condition where the sine component is zero. Specifically, this occurs when \( \sin(5\pi x) = 0 \). This happens at multiples of \( \pi \), so we set \( 5\pi x = n\pi \), which simplifies to \( x = \frac{n}{5} \).

This means that nodes occur at positions \( x = 0, \frac{1}{5}, \frac{2}{5}, \ldots \). For our exercise:
  • The smallest node location is at \( x = 0 \).
  • The second smallest node is at \( x = 0.2 \) meters.
  • The third smallest node is at \( x = 0.4 \) meters.
Wave Period
The wave period is the time it takes for a full cycle of the wave to pass a given point. For a sinusoidal standing wave, this period can be found using the cosine term of the wave function. The given function has a temporal component \( \cos(40\pi t) \), indicating that the angular frequency \( \omega = 40\pi \).

The formula to find the period \( T \) is \( T = \frac{2\pi}{\omega} \). Plugging in \( \omega = 40\pi \), we get:
  • \( T = \frac{2\pi}{40\pi} = \frac{1}{20} \) seconds.
This tells us that any non-node point on the wave pattern completes an oscillation every \( \frac{1}{20} \) seconds.
Wave Speed
Wave speed indicates how fast a wave travels through the medium. For a standing wave, the speed can be determined from its wave number and angular frequency. From our wave equation, we have \( k = 5\pi \) as the wave number and \( \omega = 40\pi \) as the angular frequency.

The wave speed \( v \) is calculated using the formula \( v = \frac{\omega}{k} \):
  • \( v = \frac{40\pi}{5\pi} = 8 \text{ meters/second} \).
This means the component waves that create the standing wave move at 8 meters per second.
Transverse Velocity
Transverse velocity refers to the velocity of a point on the string moving perpendicular to the wave's direction. For our wave, this is determined by taking the derivative of the wave function with respect to time and setting it to zero. The critical instances occur when the time-dependent factor \( \cos(40\pi t) \) is zero.

This gives us the condition \( 40\pi t = \frac{\pi}{2} + n\pi \):
  • First time all points have zero transverse velocity: \( t = \frac{1}{80} \) seconds.
  • Second time: \( t = \frac{3}{80} \) seconds.
  • Third time: \( t = \frac{1}{16} \) seconds.
These times correspond to the instances when wave movement pauses briefly before changing direction.
Wave Amplitude
Wave amplitude is the maximum displacement of points from their equilibrium position in the wave. In the context of our standing wave, the given equation suggests an amplitude of \( 0.040 \) meters for the standing wave itself.

However, the standing wave is a result of the interference of two identical traveling waves. Each of these waves individually has half the amplitude of the standing wave. Therefore, the amplitude of each traveling wave is:
  • \( 0.020 \) meters.
This measurement represents how far points on the string move above or below their rest position due to a single traveling wave component.

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Most popular questions from this chapter

One of the harmonic frequencies for a particular string under tension is \(325 \mathrm{~Hz}\). The next higher harmonic frequency is \(390 \mathrm{~Hz}\) What harmonic frequency is next higher after the harmonic frequency \(195 \mathrm{~Hz}\) ?

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string with a speed of \(10 \mathrm{~cm} / \mathrm{s}\). If the time interval between instants when the string is flat is \(0.50 \mathrm{~s}\), what is the wavelength of the waves?

A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\) with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of \(1.00 \mathrm{~cm}\). The motion is continuous and is repeated regularly 120 times per second. The string has linear density \(120 \mathrm{~g} / \mathrm{m}\) and is kept under a tension of \(90.0 \mathrm{~N}\). Find the maximum value of (a) the transverse speed \(u\) and (b) the transverse component of the tension \(\tau\). (c) Show that the two maximum values calculated above occur at the same phase values for the wave. What is the transverse displacement \(y\) of the string at these phases? (d) What is the maximum rate of energy transfer along the string? (e) What is the transverse displacement \(y\) when this maximum transfer occurs? (f) What is the minimum rate of energy transfer along the string? (g) What is the transverse displacement \(y\) when this minimum transfer occurs?

These two waves travel along the same string: \(y_{1}(x, t)=(4.60 \mathrm{~mm}) \sin (2 \pi x-400 \pi t)\) \(y_{2}(x, t)=(5.60 \mathrm{~mm}) \sin (2 \pi x-400 \pi t+0.80 \pi \mathrm{rad})\) What are (a) the amplitude and (b) the phase angle (relative to wave 1 ) of the resultant wave? (c) If a third wave of amplitude \(5.00 \mathrm{~mm}\) is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?
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