Question
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 1.3×10−4kg/m and wave equation is y=0.021sin(x+30t). Find the tension in the string where x in meter, t 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}$$Solution
The wave equation of a wave is given as y(x,t)=Asin(kx±ωt+ϕ)where,x is the position of the wave at time t, t is the time taken, A is the amplitude, k is the wavenumber and ω 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 2πrad. From the wave equation, the wave number k=λ2π and and the angular frequencyω=T2π where T is the time period of the wave and can also be written as T=f1, f is the frequency of the wave.
Formula used:
v=μT and v=λ×f
Complete step by step answer:
Given that the wave equation is y=0.021sin(x+30t)
Then, angular frequency is given as ω=30rad
Then we know that, the frequency is given asf=2πω=2π30s−1
We also know that v=λ×f where v is the velocity of the wave and λ 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
The velocity of the wave is v=λ×f=2π×2π30=30
Also, the linear density of the string is μ=1.3×10−4kg/m
We also know that v=μT where T is the tension in the string.
Then, 30=1.3×10−4T
⟹900×1.3×10−4=T
∴T=1.17×10−1N
Hence the answer is B.1.17×10−1N
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 1rad=180degree×π, which is to say that π is equivalent to 180∘.