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Question: A wave is represented by the equation \[y = 10\sin 2\pi \left( {100t - 0.02x} \right) + 10\sin 2\pi ...

A wave is represented by the equation y=10sin2π(100t0.02x)+10sin2π(100t+0.02x)y = 10\sin 2\pi \left( {100t - 0.02x} \right) + 10\sin 2\pi \left( {100t + 0.02x} \right). The maximum amplitude and loop length are respectively:
(A) 2020 units and 3030 units
(B) 2020 units and 2525 units
(C) 3030 units and 2020 units
(D) 2525 units and 2020 units

Explanation

Solution

This given problem can be solved using the concept of superposition of waves. The principle of superposition of waves enables us to determine the net waveform when any number of individual waveforms superimpose each other in the medium in which these waves are travelling.

Step-by-step solution:
Step 1: First of all, let us understand the principle of superposition of waves when the given waves are travelling in any medium.
According to superposition principle, when two or more waves are travelling through a medium superimpose each other, then a new wave is formed and the resultant displacement of the resultant wave at any instant is equal to the algebraic sum of the displacements due to individual waves at that given instant.
If two waves are given i.e., y1=10sin2π(100t0.02x)\mathop y\nolimits_1 = 10\sin 2\pi \left( {100t - 0.02x} \right) and y2=10sin2π(100t+0.02x)\mathop y\nolimits_2 = 10\sin 2\pi \left( {100t + 0.02x} \right) travelling in a medium then superposition of these two waves can be calculated as given below –
y=y1+y2y = \mathop y\nolimits_1 + \mathop y\nolimits_2
y=10sin2π(100t0.02x)+10sin2π(100t+0.02x)y = 10\sin 2\pi \left( {100t - 0.02x} \right) + 10\sin 2\pi \left( {100t + 0.02x} \right)
Now, we know that sinA+sinB=2sin(A+B2)cos(AB2)\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) where A=2π(100t0.02x)A = 2\pi \left( {100t - 0.02x} \right)and B=2π(100t+0.02x)B = 2\pi \left( {100t + 0.02x} \right)
So, using this identity in the above equation, we will get –
y=10(2sin2π(100t0.02x+100t0.02x)2cos2π(100t0.02x100t+0.02x)2)y = 10\left( {2\sin \dfrac{{2\pi \left( {100t - 0.02x + 100t - 0.02x} \right)}}{2}\cos \dfrac{{2\pi \left( {100t - 0.02x - 100t + 0.02x} \right)}}{2}} \right)
On solving this above equation –
y=20sin2π×200t2cos2π(2×0.02x)2y = 20\sin \dfrac{{2\pi \times 200t}}{2}\cos \dfrac{{2\pi \left( { - 2 \times 0.02x} \right)}}{2}, We know the identity cos(θ)=cosθ\cos ( - \theta ) = \cos \theta
y=20sin(2π×100t)cos(2π×0.02x)y = 20\sin \left( {2\pi \times 100t} \right)\cos \left( {2\pi \times 0.02x} \right) ……………….(1)
But we know the generalize from of resultant of the two superimposing waves and that is given by equation below –
y=Rsin(ωt)cos(kx)y = R\sin (\omega t)\cos (kx) …………………….(2)
Where k=2πλ=k = \dfrac{{2\pi }}{\lambda } = wave number, λ=\lambda = wavelength, R=R = resultant amplitude, and ω=2πT=\omega = \dfrac{{2\pi }}{T} = angular frequency
Step 2: Now on comparing equation (1) and (2), we will get –
Resultant amplitude i.e. R=20R = 20units
And k=2πλk = \dfrac{{2\pi }}{\lambda }, so on rearranging this λ=2πk\lambda = \dfrac{{2\pi }}{k}
Where k=2π×0.02k = 2\pi \times 0.02
So, λ=2π2π×0.02=10.02=50\lambda = \dfrac{{2\pi }}{{2\pi \times 0.02}} = \dfrac{1}{{0.02}} = 50units.
But loop length can be defined as –
λ2=502=25\dfrac{\lambda }{2} = \dfrac{{50}}{2} = 25units

So, the option (B) is the correct option.

Note:
-It should always be kept in mind that overlapping waves always add algebraically to produce the resultant wave and these overlapping waves do not alter the travel of one another in any way.
-The principle of superposition applies equally well to electro-magnetic waves. But the superposition principle ceases to apply when the amplitude of mechanical waves is too large.