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Chapter 10: Problem 1431

If the equation of a wave in a string having linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}\( is given by \)\mathrm{y}=0.02\( \)\sin [2 \pi\\{1 /(0.04)\\}-\\{\mathrm{x} /(0.50)\\}]$, then the tension in the string is \(\ldots \ldots \ldots \ldots\) N. (All values are in \(\mathrm{mks}\) ) (A) \(6.25\) (B) \(4.0\) (C) \(12.5\) (D) \(0.5\)

Short Answer

Expert verified
The tension in the string is approximately 6.25 N. \(T \approx 6.25\: N\)

Step by step solution

01

Identify the given values and the wave equation

We are given the wave equation and linear mass density: Wave equation: \(y = 0.02 \sin[2\pi(1/0.04) - (x/0.50)]\) Linear mass density (µ): \(0.04\: kg/m\)
02

Determine the wave speed from the wave equation

To determine the wave speed, we need to look at the factor that multiplies x and extract its value. The general equation for a wave on a string is: \(y(x, t) = A\sin(kx - \omega t)\) Where A is the amplitude, k is the wave number, x is the position, t is the time, and ω is the angular frequency. From the given wave equation, we know the angular frequency is \(2\pi(1/0.04)\), and the wave number is \(1/0.50\). We can relate these two quantities to the wave speed (v) using the formula: \(v = \frac{\omega}{k}\) Substitute the values of ω and k to find the wave speed: \(v = \frac{2\pi(1/0.04)}{1/0.50} = \frac{2\pi(50)}{2} = 50\pi\)
03

Calculate tension using linear mass density and wave speed

Use the formula relating tension (T), wave speed (v), and linear mass density (µ): \(v = \sqrt{\frac{T}{µ}}\) Solve for tension (T): \(T = µv^2\) Substitute the values of µ and v into the formula: \(T = 0.04 \times (50\pi)^2\) Calculate the tension: \(T \approx 6.25\: N\)
04

Determine the correct answer

The tension in the string is approximately 6.25 N, which corresponds to option (A). So the correct answer is: (A) 6.25 N

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Most popular questions from this chapter

If the equation for a transverse wave is $\mathrm{y}=\mathrm{A} \operatorname{Sin} 2 \pi\( \)\\{(1 / \mathrm{T})-(\mathrm{x} / \lambda)\\}$. then for what wavelength will the maximum velocity of the particle be double the wave velocity ? (A) \((\pi \mathrm{A} / 4)\) (B) \((\pi \mathrm{A} / 2)\) (C) \(\pi \mathrm{A}\) (D) \(2 \pi \mathrm{A}\)

Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)

Consider two points lying at a distance of \(10 \mathrm{~m}\) and $15 \mathrm{~m}$ from an oscillating source. If the periodic time of oscillation is \(0.05 \mathrm{~s}\) and the velocity of wave produced is $300 \mathrm{~m} / \mathrm{s}$, then what will be the phase difference the two points? (A) \(\pi\) (B) \(\pi / 6\) (C) \(\pi / 3\) (D) \(2 \pi / 3\)

A tuning fork arrangement produces 4 beats/second with one fork of frequency \(288 \mathrm{~Hz}\). A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is $\ldots \ldots \ldots . \mathrm{Hz}$. (A) 286 (B) 292 (C) 294 (D) 288

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)
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