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Chapter 10: Problem 1476

A string of linear density \(0.2 \mathrm{~kg} / \mathrm{m}\) is stretched with a force of \(500 \mathrm{~N}\). A transverse wave of length \(4.0 \mathrm{~m}\) and amplitude \(1 / 1\) meter is travelling along the string. The speed of the wave is \(\ldots \ldots \ldots \ldots \mathrm{m} / \mathrm{s}\) (A) 50 (B) \(62.5\) (C) 2500 (D) \(12.5\)

Short Answer

Expert verified
The speed of the wave is \(50 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Recall the formula for wave speed

We use the formula for wave speed on a string, given by: \(v = \sqrt{\frac{T}{\mu}}\) where \(v\) is the wave speed, \(T\) is the tension force, and \(\mu\) is the linear density of the string.
02

Plug in the given values

We are given the linear density \(\mu = 0.2 \mathrm{~kg} / \mathrm{m}\) and the tension force \(T = 500 \mathrm{~N}\). We can plug these values into the formula to find the wave speed: \(v = \sqrt{\frac{500 \mathrm{~N}}{0.2 \mathrm{~kg} / \mathrm{m}}}\)
03

Solve for the wave speed

Now, we can perform the calculation to find the wave speed: \(v = \sqrt{\frac{500 \mathrm{~N}}{0.2 \mathrm{~kg} / \mathrm{m}}} = \sqrt{2500 \mathrm{~m} / \mathrm{s}} = 50 \mathrm{~m} / \mathrm{s}\)
04

Compare the result with the given choices

The wave speed we calculated is \(50 \mathrm{~m} / \mathrm{s}\), which corresponds to choice (A). So the correct answer is: (A) 50

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Most popular questions from this chapter

When a mass \(\mathrm{m}\) is suspended from the free end of a massless spring having force constant \(\mathrm{k}\), its oscillates with frequency \(\mathrm{f}\). Now if the spring is divided into two equal parts and a mass $2 \mathrm{~m}$ is suspended from the end of anyone of them, it will oscillate with a frequency equal to......... (A) \(\mathrm{f}\) (B) \(2 \mathrm{f}\) (C) \((\mathrm{f} / \sqrt{2})\) (D) \(\sqrt{2 f}\)

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) $\left.\mathrm{T}=2 \pi \sqrt{\\{\ell} /\left(\mathrm{g}^{2}+\mathrm{a}^{2}\right)\right\\}$

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) \(350 \mathrm{~m} / \mathrm{s}\)

The wave number for a wave having wavelength \(0.005 \mathrm{~m}\) is \(\ldots \ldots \mathrm{m}^{-1}\) (A) 5 (B) 50 (C) 100 (D) 200
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