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प्रश्न
The displacement of an elastic wave is given by the function y = 3 sin ωt + 4 cos ωt. where y is in cm and t is in second. Calculate the resultant amplitude.
उत्तर
Given, the displacement of an elastic wave y = 3 sin ωt + 4 cos ωt
Assume, 3 = cos ϕ ......(i)
4 = a sin ϕ ......(ii)
On dividing equation (ii) by equation (i)
tan θ = `4/3`
⇒ ϕ = `tan^-1(4/3)`
Also, a2 cos2 ϕ + a2 sin2 ϕ = 32 + 42
⇒ a2 (cos2 ϕ + sin2 ϕ) = 25
a2 · 1 = 25
⇒ a = 5
Hence Υ = 5 cos ϕ sin ωt + 5 sin ϕ cos ωt
= 5[cos ϕ sin ωt + sin ϕ cos ωt]
= 5 sin(ωt + ϕ)
Where ϕ = `tan^-1 (4/3)`
Hence, amplitude = 5 cm.
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