Problem 52
Question
A \(50.0-\mathrm{cm}\) -long wire with a mass of \(10.0 \mathrm{~g}\) is under a tension of \(50.0 \mathrm{~N}\). Both ends of the wire are held rigidly while it is plucked. a) What is the speed of the waves on the wire? b) What is the fundamental frequency of the standing wave? c) What is the frequency of the third harmonic?
Step-by-Step Solution
Verified Answer
Answer: The wave speed on the wire is 50 m/s, the fundamental frequency of the standing wave is 50 Hz, and the frequency of the third harmonic is 150 Hz.
1Step 1: Calculate the linear mass density of the wire.
First, we need to calculate the linear mass density (µ) of the wire. Linear mass density is the mass per unit length of the wire and can be calculated using the formula:
µ = mass/length
where the mass is in kilograms and the length is in meters.
Mass = 10.0 g = 0.01 kg (converted from grams to kilograms)
Length = 50.0 cm = 0.5 m (converted from centimeters to meters)
µ = 0.01 kg / 0.5 m = 0.02 kg/m
2Step 2: Calculate the wave speed on the wire.
Now, we can use the formula for wave speed (v) on a string, which is given by
v = √(T/µ)
where T is the tension in the string and µ is the linear mass density.
Tension = 50.0 N
µ = 0.02 kg/m
v = √(50.0 N / 0.02 kg/m) = √(2500 m^2/s^2) = 50 m/s
The wave speed on the wire is 50 m/s.
3Step 3: Find the fundamental frequency of the standing wave.
The fundamental frequency (f1) of the standing wave can be calculated using the formula:
f1 = v / 2L
where v is the wave speed and L is the length of the wire.
v = 50 m/s
L = 0.5 m
f1 = 50 m/s / (2 * 0.5 m) = 50 Hz
The fundamental frequency of the standing wave is 50 Hz.
4Step 4: Determine the frequency of the third harmonic.
The frequency of the nth harmonic can be calculated using the formula:
fn = n * f1
where n is the harmonic number (3 in this case) and f1 is the fundamental frequency.
f3 = 3 * 50 Hz = 150 Hz
The frequency of the third harmonic is 150 Hz.
Key Concepts
Wave SpeedLinear Mass DensityHarmonicsFundamental Frequency
Wave Speed
Understanding the speed at which waves travel along a medium, such as a string or wire, is crucial in physics and various applications, including musical instruments. The wave speed (\( v \) is determined by two factors: the tension of the medium (\( T \) and its linear mass density (\( \text{mass} / \text{length} \) or µ). It can be represented mathematically by the formula: \[ v = \sqrt{\frac{T}{\mu}} \]
In this equation, a higher tension usually means faster wave speeds, since the wave can propagate more quickly when the medium resists being deformed. Conversely, a larger mass per unit length (higher linear mass density) will generally slow down the wave, as more mass requires more force to accelerate.
In this equation, a higher tension usually means faster wave speeds, since the wave can propagate more quickly when the medium resists being deformed. Conversely, a larger mass per unit length (higher linear mass density) will generally slow down the wave, as more mass requires more force to accelerate.
Linear Mass Density
Linear mass density, denoted as µ, is a measure of how much mass is distributed along the length of a string or wire. It's a fundamental property that influences how waves behave on that medium. \[ \mu = \frac{\text{mass}}{\text{length}} \]
For example, a thicker guitar string has a greater linear mass density than a thinner one, which affects the pitch of the notes it produces when plucked. Calculating µ is the first step in determining wave characteristics and is crucial for ensuring that our calculations for wave speed and subsequent frequencies are accurate.
For example, a thicker guitar string has a greater linear mass density than a thinner one, which affects the pitch of the notes it produces when plucked. Calculating µ is the first step in determining wave characteristics and is crucial for ensuring that our calculations for wave speed and subsequent frequencies are accurate.
Harmonics
Harmonics are integral to understanding musical tones and several physical phenomena. They are essentially multiples of the fundamental frequency, which create the diverse sounds we hear from musical instruments. In a plucked string, the first harmonic is the fundamental frequency; subsequent harmonics (like the third harmonic mentioned in the exercise) are integer multiples of this base pitch. Each harmonic corresponds to a standing wave pattern with more antinodes – points of maximum amplitude – and nodes – points of zero amplitude.
Fundamental Frequency
The fundamental frequency is the lowest frequency of a periodic waveform and is associated with the first harmonic of a system. It sets the pitch of the note and is determined by the properties of the medium—in our case, the wire. The string's tension, length, and mass all play a part in this base frequency.
Mathematically, it's calculated with the formula: \[ f_1 = \frac{v}{2L} \] where \( f_1 \) is the fundamental frequency, \( v \) is the wave speed, and \( L \) is the length of the wire. This lowest-frequency standing wave has nodes only at the ends and one antinode in the middle, corresponding to the loudest part of the wave.
Mathematically, it's calculated with the formula: \[ f_1 = \frac{v}{2L} \] where \( f_1 \) is the fundamental frequency, \( v \) is the wave speed, and \( L \) is the length of the wire. This lowest-frequency standing wave has nodes only at the ends and one antinode in the middle, corresponding to the loudest part of the wave.
Other exercises in this chapter
Problem 49
A guitar string with a mass of \(10.0 \mathrm{~g}\) is \(1.00 \mathrm{~m}\) long and attached to the guitar at two points separated by \(65.0 \mathrm{~cm} .\) a
View solution Problem 50
Write the equation for a sinusoidal wave propagating in the negative \(x\) -direction with a speed of \(120 . \mathrm{m} / \mathrm{s}\), if a particle in the me
View solution Problem 53
What is the wave speed along a brass wire with a radius of \(0.500 \mathrm{~mm}\) stretched at a tension of \(125 \mathrm{~N}\) ? The density of brass is \(8.60
View solution Problem 54
Two steel wires are stretched under the same tension. The first wire has a diameter of \(0.500 \mathrm{~mm}\), and the second wire has a diameter of \(1.00 \mat
View solution