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Chapter 16: Problem 73

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of \(45 \mathrm{~Hz}\). The mass of the wire is \(3.5 \times 10^{-2} \mathrm{~kg}\) and its linear mass density is \(4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\). What is (i) the speed of a transverse wave on the string and (ii) the tension in the string? (a) (i) \(80 \mathrm{~m} / \mathrm{s}\) (ii) \(250 \mathrm{~N}\) (b) (i) \(88 \mathrm{~m} / \mathrm{s}\) (ii) \(208 \mathrm{~N}\) (c) (i) \(90 \mathrm{~m} / \mathrm{s}\) (ii) \(249 \mathrm{~N}\) (d) (i) \(78.75 \mathrm{~m} / \mathrm{s}\) (ii) \(248 \mathrm{~N}\)

Short Answer

Expert verified
Option (d) is correct.

Step by step solution

01

Determine the length of the wire

To find the length of the wire, divide the total mass of the wire by its linear mass density. Given mass = \(3.5 \times 10^{-2} \mathrm{~kg}\) and linear mass density \(\mu = 4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\), we calculate the length, \(L\).\[ L = \frac{\text{mass}}{\text{linear mass density}} = \frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}} = 0.875 \mathrm{~m} \]
02

Calculate the speed of the transverse wave

The speed of the wave, \(v\), in the fundamental mode is given by the formula \(v = 2fL\), where \(f = 45 \mathrm{~Hz}\) is the frequency and \(L = 0.875 \mathrm{~m}\). Substitute the values:\[ v = 2 \times 45 \times 0.875 = 78.75 \mathrm{~m/s} \]
03

Calculate the tension in the string

The tension, \(T\), in the string can be computed using the formula \(v = \sqrt{\frac{T}{\mu}}\). Rearranging gives: \(T = \mu v^2\). Use \(\mu = 4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\) and \(v = 78.75 \mathrm{~m/s}\):\[ T = 4.0 \times 10^{-2} \times (78.75)^2 = 248 \mathrm{~N} \]
04

Choose the correct answer

The calculated values match option (d): (i) \(78.75 \mathrm{~m/s}\) and (ii) \(248 \mathrm{~N}\). Thus, option (d) is the correct choice.

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Key Concepts

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

Wave Speed Calculation
Calculating the speed of a wave on a stretched string is quite interesting. Here's how it goes. Wave speed, denoted as \(v\), is essentially the speed at which disturbances travel along the medium. The formula to find this speed while the wave vibrates in its fundamental mode is \(v = 2fL\). Here:
  • \(f\) is the frequency of the wave. For example, in this exercise, it's given as \(45\, \text{Hz}\).
  • \(L\) is the length of the string. It's calculated by dividing the total mass by the linear mass density, which determines how much mass the string has over a length unit.
To use the formula properly, multiply the frequency by two, then by the length. That gives you the wave speed. With our numbers, you will end up with a wave traveling at a speed of \(78.75\, \text{m/s}\). Simple, right? Just like learning to ride a bike, once you get the hang of it, calculation becomes straightforward!
String Tension
The tension in the string is another exciting aspect of waves. It's like the pull that makes the string taut and allows waves to travel. We find it with the formula \(v = \sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the linear mass density of the string.
To solve for tension, you rearrange the formula to \(T = \mu v^2\). In our example, you substitute \(\mu = 4.0 \times 10^{-2} \, \text{kg/m}\) and \(v = 78.75 \, \text{m/s}\). Completing the math ends up with a tension of \(248 \, \text{N}\).
  • The linear mass density tells you how heavy the string is evenly over its length.
  • The tension increases with the square of the wave speed, meaning a higher speed dramatically affects tension.
Understanding how to calculate tension helps in comprehending the forces at play on a string. Like understanding how tight to pull a guitar string for the right note!
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system oscillates, and for strings, it's pretty important. In our string, it鈥檚 the frequency that allows the simplest possible wave - one loop.
When a string is plucked, it may produce various harmonics, but the fundamental frequency is the basis 鈥� given as \(45\,\text{Hz}\) in the exercise.
  • It depends on length, mass per unit length (linear density), and tension in the string.
  • Higher tension produces higher frequency, and a heavier string lowers pitch.
It's like tuning into your favorite radio station, where only one station can be tuned to at a time - that's the easiest natural vibration the string wants to hum to.
That鈥檚 why understanding fundamental frequency is vital for realistic sound production in musical instruments and for solving wave equations effectively.

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Most popular questions from this chapter

Two identical sounds \(A\) and \(B\) reach a point in the same phase. The resultant sound is \(C\). The loudness of \(C\) is \(n \mathrm{~dB}\) higher than the loudness of \(A\). (a) 2 (b) 3 (c) 4 (d) 6

Two uniform strings \(A\) and \(B\) made of steel are made to vibrate under the same tension. If the first overtone of \(A\) is equal to the second overtone of \(B\) and if the radius of \(A\) is twice that of \(B\), the ratio of the lengths of the strings is (a) \(2: 1\) (b) \(3: 4\) (c) \(3: 2\) (d) \(1: 3\)

A source and an observer move away from each other with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) with respect to ground. If the observer finds the frequency of sound coming from the source as \(1950 \mathrm{~Hz}\), then actual frequency of the source is (velocity of sound in air \(=340 \mathrm{~m} / \mathrm{s}\) ) (a) \(1950 \mathrm{~Hz}\) (b) \(2068 \mathrm{~Hz}\) (c) \(2132 \mathrm{~Hz}\) (d) \(2486 \mathrm{~Hz}\)

The path difference between two waves \(y_{1}=a_{1} \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)\) and \(y_{2}=a_{2} \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)\) is (a) \(\frac{\lambda}{2 \pi}(\phi)\) (b) \(\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)\) (c) \(\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)\) (d) \(\frac{2 \pi}{\lambda}(\phi)\)

When 2 tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats \(\mathrm{s}^{-1}\) are heard. Now, some tape is attached on the prong of fork \(2 .\) When the tuning forks are sounded again, 6 beats \(\mathrm{s}^{-1}\) are heard if the frequency of fork 1 is \(200 \mathrm{~Hz}\), then what was the original frequency of fork 2 ? (a) \(196 \mathrm{~Hz}\) (b) \(200 \mathrm{~Hz}\) (c) \(202 \mathrm{~Hz}\) (d) \(204 \mathrm{~Hz}\)
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