1.

The displacement of an elastic wave is given the function y = 3 sin ωt + 4 cos ωt.When y is in cm and t is in second. Calculate the resultant amplitude.

Answer»

y =3sin ωt +4 cos ωt …(i)

Let, 3 = a cos ϕ …(ii)

4 = a sin ϕ …(iii)

∴ = a cos ϕ sin ωt + a sinϕ cos ωt …(iv)

y = a sin(ωt + φ)

From eq. (s) (i) & (iii)-

tan ϕ = \(\frac{3}{4}\) or ϕ = tan-1\(\frac{3}{4}\) 

Squaring (ii) & (iii), then adding- 

a2cos2ϕ+ a2sin2ϕ = 32 + 42

a2(cos2ϕ + sin2ϕ) = 9 + 16

a2 = 25

or a = 5 or amplitude, = 5 cm

∴ y1 = 5 sin (ωt + ϕ), ∵ ϕ = tan-1\(\frac{4}{3}\) 

Hence, new amplitude, a = 5cm.



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