91Ó°ÊÓ

Suppose that depicts standing waves on a metal string under a tension of \(88.2 \mathrm{~N}\). Its length is \(50.0 \mathrm{~cm}\) and its mass is \(0.500\) g. ( \(a\) ) Compute \(v\) for transverse waves on the string. (b) Determine the frequencies of its fundamental, first overtone, and second overtone. (a) \(v=\sqrt{\frac{\text { Tension }}{\sqrt{\text { Mass per t unit length }}}}=\sqrt{\frac{88.2 \mathrm{~N}}{15.00 \times 10^{-4} \mathrm{~kg} /(0.500 \mathrm{~m})}}=297 \mathrm{~m} / \mathrm{s}\) (b) We recall that the length of the segment is \(\lambda / 2\) and we use \(\lambda=\) \(\mathrm{u} / \mathrm{f}\). For the fundamental: $$ \lambda=1.00 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{1.00 \mathrm{~m}}=297 \mathrm{~Hz} $$ For the first overtone: $$ \lambda=0.500 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.500 \mathrm{~m}}=594 \mathrm{~Hz} $$ For the second overtone: $$ \lambda=0.333 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.333 \mathrm{~m}}=891 \mathrm{~Hz} $$

Step by step solution

01

Calculate the mass per unit length

Convert the mass of the string from grams to kilograms: Mass = 0.500 g = 0.0005 kgFind the mass per unit length (\( \mu \)):\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{0.0005}{0.50} = 0.001 \text{ kg/m} \]

02

Calculate the wave speed

Use the formula for wave speed on a string: \[ v = \sqrt{\frac{\text{Tension}}{\mu}} \]Given the tension is 88.2 N, calculate \( v \):\[ v = \sqrt{\frac{88.2}{0.001}} = \sqrt{88200} = 297 \text{ m/s} \]

03

Find the fundamental frequency

The fundamental frequency, \( f_1 \), corresponds to the string supporting half a wavelength:\( \lambda = 2L = 1.0 \text{ m} \).Calculate the frequency:\[ f_1 = \frac{v}{\lambda} = \frac{297}{1.0} = 297 \text{ Hz} \]

04

Calculate the first overtone

The first overtone corresponds to a full wavelength (\( \lambda = L = 0.5 \text{ m} \)).Calculate the frequency:\[ f_2 = \frac{v}{\lambda} = \frac{297}{0.5} = 594 \text{ Hz} \]

05

Calculate the second overtone

The second overtone corresponds to 1.5 wavelengths (\( \lambda = \frac{2}{3}L = 0.333 \text{ m} \)).Calculate the frequency:\[ f_3 = \frac{v}{\lambda} = \frac{297}{0.333} = 891 \text{ Hz} \]

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Key Concepts

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

Wave Speed Calculation

Understanding how wave speed is calculated on a stretched string is essential for grasping wave dynamics. Wave speed (\(v\)) on a string is determined by two primary factors: the tension in the string and the mass per unit length (\(\mu\)). The formula used to calculate wave speed is:
\[v = \sqrt{\frac{\text{Tension}}{\mu}}\]This equation tells us that the wave speed increases with greater tension and decreases with a higher mass per unit length.
### Practical ExampleIn our problem, we calculated wave speed on a string with a tension of 88.2 N. First, we needed to find the mass per unit length (\(\mu = 0.001 \text{ kg/m}\)). Substituting these values into the formula, we got:

• Wave speed, \(v = \sqrt{\frac{88.2}{0.001}} = \sqrt{88200} = 297 \text{ m/s}\)

This speed helps determine how fast the waves can travel along the string, which is crucial for finding different frequencies of standing waves.

Harmonics and Overtones

When waves oscillate on a string, they create standing waves which consist of nodes and antinodes. These standing waves can vibrate at specific frequencies called harmonics. The fundamental frequency is the simplest form of vibration, while higher frequencies are known as overtones.
### Understanding Harmonics- **Fundamental frequency**: The simplest and lowest frequency where the string vibrates, corresponding to one-half of a wavelength.- **First overtone**: Occurs when the string vibrates at twice the frequency of the fundamental; it has one full wavelength.- **Second overtone**: Occurs at three times the frequency of the fundamental and corresponds to 1.5 wavelengths.
### Calculating the FrequenciesFrom the original problem:

• Fundamental frequency, \(f_1 = 297 \text{ Hz}\) when \(\lambda = 1.0 \text{ m}\).
• First overtone, \(f_2 = 594 \text{ Hz}\) with \(\lambda = 0.5 \text{ m}\).
• Second overtone, \(f_3 = 891 \text{ Hz}\), for \(\lambda = 0.333 \text{ m}\).

Each overtone represents a new standing wave pattern on the string, increasing in frequency due to shorter wavelengths.

Mass Per Unit Length

Mass per unit length (\(\mu\)) is a critical parameter for determining the wave speed on a string. It is defined as the mass of the string divided by its length. A string with more mass per unit length tends to slow down wave propagation.
### CalculationIn our exercise, we have:
- **Mass of the string**: 0.500 grams. This was converted to kilograms for consistent units, giving us 0.0005 kg.- **Length of the string**: 50.0 cm, converted to meters gives 0.5 m.
Thus, mass per unit length is calculated as:

• \(\mu = \frac{0.0005}{0.5} = 0.001 \text{ kg/m}\)

This parameter influences how fast waves propagate along the string. A higher mass per unit length means a lower wave speed at the same tension, impacting the frequencies of the standing waves.