Standing waves are produced by the superposition of two waves with y1=0.05 sin find the amplitude of particle 0.5m
Answers
The amplitude of particle at 0.5m is 5.4 × 10⁻²m.
Explanation:
Q: Standing waves are produced by the superposition of two waves y1 = 0.05 sin(3πt – 2x) and y2 = 0.05 sin(3πt + 2x) where x and y are in meters and t is in seconds. The amplitude of a particle at x = 0.5 m is………
=> Resultant displacement of waves = y1 + y2
= 0.05 sin ( 3πt – 2x ) + 0.05 sin ( 3πt + 2x )
= 0.05 [ sin ( 3πt – 2x ) + sin ( 3πt + 2x ) ]
y = 0.05 * 2 sin (3πt) cos {(– 4x)/2} ....1)
As sin R + sin S = 2sin [(R + S)/2} cos{(R – S)/2)]
from equation (1) we can say,
y = 0.1 cos2x ∙ sin3πt
=> Now, The amplitude of a particle at x = 0.5 m:
y = A sin3πt
where,
Amplitude, A = 0.1 cos2x
= 0.1 cos(2 × 0.5)
= 0.1 cos (1 rad)
=> But one radian is equals to 180/π degrees.
∴ A = 0.1 cos {(1 × 180) / π}°
= 0.1 cos ( 180/3.14 )°
= 0.1 cos (57.4)°
But, cos 57.4° is equals to 0.5387.
∴ A = 0.1 * 0.54
= 0.054 m
= 5.4 × 10⁻² m
Thus, the amplitude of particle at 0.5m is 5.4 × 10⁻² metre.
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