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Chapter 15: Problem 50

A 1750-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (\(A\) and \(B\)), each 1.25 m long and weighing 0.290 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? Which pulse arrives first? (Ignore the effect of the weight of the wires on the tension in the wires.)

Short Answer

Expert verified
The pulse through wire B arrives 0.00234 s before the pulse through wire A.

Step by step solution

01

Identify the problem

We need to find the time delay between pulses transmitted to the ceiling through the two wires, A and B, when plucked at the same time. We know the mass of the beam and the length and weight of wires.
02

Determine the position of the center of gravity

The center of gravity is one-third the way from one end (wire A) along the beam, which implies that two-thirds of the beam's weight contributes to wire B and one-third to wire A.
03

Calculate tension in each wire

Using the beam's total weight (1750 N), the tension in wire A \[ T_A = rac{1}{3} imes 1750 ext{ N} = 583.33 ext{ N} \]And the tension in wire B \[ T_B = rac{2}{3} imes 1750 ext{ N} = 1166.67 ext{ N} \]
04

Determine wave speed in each wire

The speed of wave (\( v \)) in a wire is determined by\[ v = \sqrt{\frac{T}{\mu}} \]where\( \mu = \frac{m}{L} \)is the mass per unit length. Using the weight of the wire (0.290 N) to find its mass,\( m = 0.290 ext{ N} / 9.8 ext{ m/s}^2 = 0.0296 ext{ kg} \).Mass per unit length\( \mu = 0.0296 ext{ kg} / 1.25 ext{ m} = 0.02368 ext{ kg/m} \).Then calculate:\[ v_A = \sqrt{\frac{583.33}{0.02368}} \approx 156.7 ext{ m/s} \]\[ v_B = \sqrt{\frac{1166.67}{0.02368}} \approx 221.5 ext{ m/s} \]
05

Determine arrival time of waves

Using the speeds, the time taken for the wave to travel each wire is\[ t_A = \frac{1.25 ext{ m}}{156.7 ext{ m/s}} \approx 0.00798 ext{ s} \]\[ t_B = \frac{1.25 ext{ m}}{221.5 ext{ m/s}} \approx 0.00564 ext{ s} \]
06

Calculate the time delay and determine which pulse arrives first

The time delay is given by the difference in arrival times,\[ \Delta t = t_A - t_B \approx 0.00798 ext{ s} - 0.00564 ext{ s} = 0.00234 ext{ s} \].The pulse through wire B arrives first.

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Key Concepts

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

Tension in Wires
Tension in wires plays a crucial role in wave propagation. When a beam is suspended by wires, the tension is determined by the weight distributed on each wire. In our example, the beam weighs 1750 N with the center of gravity located one-third from wire A.
This means two-thirds of the weight is supported by wire B, resulting in varying tensions:
  • For wire A, the tension is calculated as one-third of the beam's weight: \( T_A = \frac{1}{3} \times 1750 \text{ N} = 583.33 \text{ N} \).
  • For wire B, the tension is two-thirds of the beam's weight: \( T_B = \frac{2}{3} \times 1750 \text{ N} = 1166.67 \text{ N} \).
Higher tension in a wire allows faster wave propagation. This principle is essential in engineering and physics applications where tension significantly affects the behavior of structures and transmitted waves.
Center of Gravity
The center of gravity refers to the point at which an object's mass is evenly distributed. In the context of a suspended beam, it affects how weight is shared between supporting wires.
In our scenario, the center of gravity is one-third of the distance from the end where wire A is attached. This placement indicates:
  • One-third of the beam's weight is supported by wire A.
  • Two-thirds of the beam's weight is supported by wire B.
Understanding the center of gravity is vital in determining the force distribution across a structure, influencing engineering designs for stability and balance. It is crucial for ensuring structures like bridges and buildings can withstand applied forces without tipping.
Mass per Unit Length
Mass per unit length ( \( \mu \) ) represents the mass distributed over a given length of material. It is critical in calculating wave speed in wires. To find \( \mu \) , use the formula:\[ \mu = \frac{m}{L} \] where \( m \) is the mass and \( L \) the length of the wire.In our example, the wire weighs 0.290 N, which gives it a mass of:
  • \( m = \frac{0.290 \text{ N}}{9.8 \text{ m/s}^2} = 0.0296 \text{ kg} \).
  • With a length of 1.25 m, the mass per unit length is: \( \mu = \frac{0.0296 \text{ kg}}{1.25 \text{ m}} = 0.02368 \text{ kg/m} \).
The smaller the mass per unit length, the higher the wave speed, illustrating why lighter, longer wires are effective in faster wave transmission. This concept is widely applied in crafting musical instruments, enhancing designs for better acoustics.

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Most popular questions from this chapter

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 Hz to about 20.0 kHz. (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short- wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 m/s. How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

A sinusoidal transverse wave travels on a string. The string has length 8.00 m and mass 6.00 g. The wave speed is 30.0 m/s, and the wavelength is 0.200 m. (a) If the wave is to have an average power of 50.0 W, what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda\)/4, and (iii) \(x = \lambda\)8, from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

At a distance of \(7.00 \times 10^{12}\) m from a star, the intensity of the radiation from the star is 15.4 W/m\(^2\). Assuming that the star radiates uniformly in all directions, what is the total power output of the star?
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