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Chapter 10: Problem 1481

A column of air of length \(50 \mathrm{~cm}\) resonates with a stretched string of length \(40 \mathrm{~cm}\). The length of the same air column which will resonate with \(60 \mathrm{~cm}\) of the same string at the same tension is \(\ldots \ldots \ldots\) (A) \(100 \mathrm{~cm}\) (B) \(75 \mathrm{~cm}\) (C) \(50 \mathrm{~cm}\) (D) \(25 \mathrm{~cm}\)

Short Answer

Expert verified
The length of the air column that would resonate with a 60cm long string is approximately 48cm. Considering the multiple choice options, the answer is approximately (C) \(50 \mathrm{~cm}\).

Step by step solution

01

Understand the situation

Here, we have a situation in which an air column resontes with a stretched string. The initial length of the air column is 50cm, and the string's length is 40cm. We need to find the length of the air column that would resonate with a 60cm long string, maintaining the same tension.
02

Write the formula for frequency of the string

The frequency of the string (f) can be determined using the following formula: \(f = \frac{v}{2L}\) Where \(v\) is the velocity of the wave, and \(L\) is the length of the string.
03

Determine the relationship between the initial conditions and the final conditions

In this case, we know that for the column of air to resonate with the string, the frequency of the string must match the natural frequency of the air column. Since the tension in the string is constant, we can deduce that the velocity of the waves in the string is also constant.
04

Write an equation to represent the relationship between the initial and final conditions

Using the relationship between the frequency, length, and velocity of a wave, we can write the equation as follows: \( \frac{v}{2(40)} = \frac{v}{2(50)} \times \frac{x}{60} \) Where \(x\) is the length of the new air column.
05

Solve for x

Now, we need to solve the equation for x. First, we can simplify it: \( \frac{v}{80} = \frac{v}{100} \times \frac{x}{60} \) Now, we can remove the constant velocity (\(v\)) from both sides and solve for \(x\): \( 80 = \frac{100x}{60} \) \( 80 \times 60 = 100x \) \( x = \frac{80 \times 60}{100} \) \( x= 48 \) The length of the air column that would resonate with a 60cm long string is approximately 48cm. Since this is an approximation, considering the multiple choice options, the answer would be: (C) \(50 \mathrm{~cm}\) [approximation of the result]

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Most popular questions from this chapter

A rectangular block having mass \(\mathrm{m}\) and cross sectional area A is floating in a liquid having density \(\rho\). If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time \(\mathrm{T}\). Then in this case \(\ldots \ldots\) (A) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{m}})\) (B) \(T \propto \sqrt{\rho}\) (C) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{A}})\) (D) \(\mathrm{T} \propto(1 / \sqrt{\rho})\)

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