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Question: Linear density of a string \(1.3\times 10^{-4}kg/m\) and wave equation is \(y=0.021 sin(x+30t)\). Fi...

Linear density of a string 1.3×104kg/m1.3\times 10^{-4}kg/m and wave equation is y=0.021sin(x+30t)y=0.021 sin(x+30t). Find the tension in the string where xx in meter, tt in sec.

& A.1.17\times {{10}^{-2}}N \\\ & B.1.17\times {{10}^{-1}}N \\\ \ & C.1.17\times {{10}^{-3}}N \\\ & D.\text{ }none \\\ \end{aligned}$$
Explanation

Solution

The wave equation of a wave is given as y(x,t)=Asin(kx±ωt+ϕ)y(x,t)=Asin(kx\pm\omega t+\phi)where,xx is the position of the wave at time tt, tt is the time taken, AA is the amplitude, kk is the wavenumber and ω\omega is the angular frequency of the wave. This is used to describe the motion of the wave. Generally, one wavelength is the phase difference of 2πrad2\pi rad. From the wave equation, the wave number k=2πλk=\dfrac{2\pi}{\lambda} and and the angular frequencyω=2πT\omega=\dfrac{2\pi}{T} where TT is the time period of the wave and can also be written as T=1fT=\dfrac{1}{f}, ff is the frequency of the wave.

Formula used:
v=Tμv=\sqrt{\dfrac{T}{\mu}} and v=λ×fv=\lambda\times f

Complete step by step answer:
Given that the wave equation is y=0.021sin(x+30t)y=0.021 sin(x+30t)
Then, angular frequency is given as ω=30rad\omega=30 rad
Then we know that, the frequency is given asf=ω2π=302πs1f=\dfrac{\omega}{2\pi}=\dfrac{30}{2\pi }s^{-1}
We also know that v=λ×fv=\lambda\times f where vv is the velocity of the wave and λ\lambda is the wavelength of the wave.
Generally, the wave equation is written for one wavelength, then wavelength of the wave is given as λ=2πrad\lambda=2\pi rad

The velocity of the wave is v=λ×f=2π×302π=30v=\lambda\times f=2\pi\times\dfrac{30}{2\pi}=30
Also, the linear density of the string is μ=1.3×104kg/m\mu=1.3\times 10^{-4}kg/m
We also know that v=Tμv=\sqrt{\dfrac{T}{\mu}} where TT is the tension in the string.
Then, 30=T1.3×10430=\sqrt{\dfrac{T}{1.3\times 10^{-4}}}
    900×1.3×104=T\implies 900\times 1.3\times 10^{-4}=T
T=1.17×101N\therefore T=1.17\times 10^{-1} N
Hence the answer is B.1.17×101NB.1.17\times {{10}^{-1}}N

Note:
Waves are sinusoidal in nature and are generally expressed in terms of sine functions. But they can also be expressed as cosine functions. We know that the angles can be represented in the terms of degrees or radians. Here in wave functions, we generally use radians. Also, we can convert radians to degree and vice versa using 1  rad=degree  ×π1801 \;rad=\dfrac{degree\;\times\pi}{180}, which is to say that π\pi is equivalent to 180{{180}^{\circ }}.