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A sonometer wire supports a \(4 \mathrm{~kg}\) load and vibrates in fundamental mode with a tuning fork of frequency \(416 \mathrm{~Hz}\). The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to \(\ldots \ldots \ldots\) (A) \(1 \mathrm{~kg}\) (B) \(2 \mathrm{~kg}\) (C) \(8 \mathrm{~kg}\) (D) \(16 \mathrm{~kg}\)

Short Answer

Expert verified
The new load required to maintain the fundamental mode when the length of the wire between the bridges is doubled is \(m' = 8 \mathrm{~kg}\). So, the correct answer is (C) \(8 \mathrm{~kg}\).

Step by step solution

01

Write down the known values.

The given values are: - Load: \(m = 4 \mathrm{~kg}\) - Frequency: \(f = 416 \mathrm{~Hz}\) - Length of wire between bridges is doubled
02

Understand the relationship between load, frequency, and length of a sonometer wire.

The fundamental frequency of a sonometer wire depends on its tension, linear density, and length. This can be expressed by the formula: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \) where: - \(f\) - fundamental frequency - \(L\) - length of the wire - \(T\) - tension in the wire - \(\mu\) - linear density (mass per unit length) of the wire
03

Write down the equation for load in terms of the given values.

The tension in the wire \(T\) can be expressed in terms of the load since tension is just the force acting on the wire due to the load. The equation for tension is given by: \( T = mg \) where: - \(m\) - load - \(g\) - acceleration due to gravity
04

Incorporate the tension equation into the fundamental frequency formula.

Substitute the tension in terms of load \(m\): \( f = \frac{1}{2L} \sqrt{\frac{mg}{\mu}} \)
05

Consider the doubling of the length of the wire between bridges.

When the length of the wire between bridges is doubled, it is given that fundamental mode must be maintained. Therefore, the new frequency (\(f'\)) is the same as the original frequency: \( f' = f \) For the new length (\(2L\)) and new load (\(m'\)), the equation for frequency will be: \( f' = \frac{1}{4L} \sqrt{\frac{m'g}{\mu}} \) Both the original and new frequencies are the same, so we set the equations equal to each other: \( \frac{1}{2L} \sqrt{\frac{mg}{\mu}} = \frac{1}{4L} \sqrt{\frac{m'g}{\mu}} \)
06

Solve for the new load (\(m'\)).

To solve for \(m'\), we can cancel the common terms on both sides of the equation: \( \sqrt{\frac{m}{2}} = \sqrt{\frac{m'}{4}} \) Squaring both sides will give: \( \frac{m}{2} = \frac{m'}{4} \) and solve for the new load \(m'\): \( m' = 2m = 2 \times 4 \mathrm{~kg} \)
07

Determine the correct answer.

The new load required to maintain the fundamental mode when the length of the wire between the bridges is doubled is: \( m' = 8 \mathrm{~kg} \) So, the correct answer is (C) \(8 \mathrm{~kg}\).

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Key Concepts

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

Fundamental Frequency
In the world of musical instruments and physics experiments, understanding fundamental frequency is key. This frequency is the lowest frequency at which a system vibrates naturally. For a sonometer wire, the fundamental frequency is the main vibration heard or measured when it is excited. It plays a crucial role in determining the pitch of sound produced by stringed instruments.

The formula for calculating the fundamental frequency of a sonometer wire is:
  • \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)
Here:
  • \( f \) is the fundamental frequency.
  • \( L \) is the length of the wire.
  • \( T \) represents the tension in the wire.
  • \( \mu \) is the linear density, or mass per unit length of the wire.
Understanding this equation helps us see how changes in wire length, tension, and linear density affect the fundamental frequency. If you increase the wire's length, the frequency decreases, resulting in a lower pitch. This concept is essential in various fields such as music, engineering, and acoustics.
Tension in a Wire
Tension in a wire is the force exerted along the length of the wire, keeping it taut. In the sonometer wire setup, tension is produced by the weight (load) attached to it. The relationship between load and tension is straightforward and is given by the equation:
  • \( T = mg \)
Where:
  • \( T \) stands for tension in the wire.
  • \( m \) is the mass of the load (in kg).
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \mathrm{~m/s^2} \).
When working with a sonometer, altering the tension will change the pitch of the sound produced. Therefore, understanding and controlling the tension is vital for achieving desired frequencies, especially for stringed musical instruments and experimental physics applications.
Linear Density
Linear density is a pivotal concept when analyzing vibrating wires. Defined as the mass per unit length of the wire, it is denoted by \( \mu \) and influences the wire's vibration characteristics. The formula for linear density is:
  • \( \mu = \frac{m_{wire}}{L} \)
Where:
  • \( \mu \) is the linear density.
  • \( m_{wire} \) is the mass of the wire.
  • \( L \) is the length of the wire.
A higher linear density implies more mass for a given length, affecting the wire's ability to vibrate at higher frequencies. It is an essential factor in applications that require precise control over frequency, such as music and communication technologies.
Vibrating String
When it comes to sound production, vibrating strings are fundamental. The sound from stringed instruments, like guitars or violins, arises from vibrating strings. In a sonometer, the wire acts as a string, vibrating to produce sound waves depending on the tension, length, and material properties of the wire.

As the wire vibrates, it creates standing waves with nodes and antinodes along its length. The fundamental frequency is often the most prominent sound wave, with overtones or harmonics being additional frequencies produced. These concepts are not only pertinent in musical contexts but also in physics to study wave behavior and resonance phenomena.

Thus, knowing how to manipulate the wire's length and tension can control the vibration, and consequently, the sound it makes. This understanding is key in instrument design and tuning, acoustic analysis, and even crafting audio technology.

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Most popular questions from this chapter

A rocket is moving at a speed of \(130 \mathrm{~m} / \mathrm{s}\) towards a stationary target. While moving, it emits a wave of frequency \(800 \mathrm{~Hz}\). Calculate the frequency of the sound as detected by the target. (Speed of wave \(=330 \mathrm{~m} / \mathrm{s}\) ) (A) \(1320 \mathrm{~Hz}\) (B) \(2540 \mathrm{~Hz}\) (C) \(1270 \mathrm{~Hz}\) (D) \(660 \mathrm{~Hz}\)

If the maximum frequency of a sound wave at room temperature is \(20,000 \mathrm{~Hz}\) then its minimum wavelength will be approximately \(\ldots \ldots\left(\mathrm{v}=340 \mathrm{~ms}^{-1}\right)\) (A) \(0.2 \AA\) (B) \(5 \AA\) (C) \(5 \mathrm{~cm}\) to \(2 \mathrm{~m}\) (D) \(20 \mathrm{~mm}\)

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

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Equation for a harmonic progressive wave is given by \(\mathrm{y}=\mathrm{A}\) \(\sin (15 \pi t+10 \pi x+\pi / 3)\) where \(x\) is in meter and \(t\) is in seconds. This wave is \(\ldots \ldots\) (A) Travelling along the positive \(\mathrm{x}\) direction with a speed of \(1.5 \mathrm{~ms}^{-1}\) (B) Travelling along the negative \(\mathrm{x}\) direction with a speed of \(1.5 \mathrm{~ms}^{-1} .\) (C) Has a wavelength of \(1.5 \mathrm{~m}\) along the \(-\mathrm{x}\) direction. (D) Has a wavelength of \(1.5 \mathrm{~m}\) along the positive \(\mathrm{x}\) - direction.

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