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Chapter 10: Problem 1488

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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Most popular questions from this chapter

As shown in figure, a block A having mass \(M\) is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ' \(\mathrm{m}\) ' is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block's is \(\mu\). (A) \(A_{\max }=\\{(\mu \mathrm{mg}) / \mathrm{k}\\}\) (B) \(A_{\max }=[\\{\mu(m+M) g\\} / k]\) (C) \(A_{\max }=[\\{\mu(M-\mathrm{m}) g\\} / \mathrm{k}]\) (D) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)

A simple pendulum having length \(\ell\) is suspended at the roof of a train moving with constant acceleration 'a' along horizontal direction. The periodic time of this pendulum is.... (A) \(\mathrm{T}=2 \pi \sqrt{(\ell / \mathrm{g})}\) (B) \(\mathrm{T}=2 \pi \sqrt{\\{\ell /(\mathrm{g}+\mathrm{a})\\}}\) (C) \(\mathrm{T}=2 \pi \sqrt{\\{\ell /(\mathrm{g}-\mathrm{a})\\}}\) (D) $\left.\mathrm{T}=2 \pi \sqrt{\\{\ell} /\left(\mathrm{g}^{2}+\mathrm{a}^{2}\right)\right\\}$

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

If the mass of 1 mole of air is \(29 \times 10^{-3} \mathrm{~kg}\), then the speed of sound in it at STP is $(\gamma=7 / 5) .\left\\{\mathrm{T}=273 \mathrm{~K}, \mathrm{P}=1.01 \times 10^{5} \mathrm{~Pa}\right\\}$ (A) \(270 \mathrm{~m} / \mathrm{s}\) (B) \(290 \mathrm{~m} / \mathrm{s}\) (C) \(330 \mathrm{~m} / \mathrm{s}\) (D) \(350 \mathrm{~m} / \mathrm{s}\)
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