Standing waves are produced by the superposition of two waves `y_(1)=0.05 sin (3 pit-2x) and y_(2) = 0.05 sin (3pit+2x)` Where x and y are in metres a

Question

Standing waves are produced by the superposition of two waves `y_(1)=0.05 sin (3 pit-2x) and y_(2) = 0.05 sin (3pit+2x)` Where x and y are in metres a

Standing waves are produced by the superposition of
two waves
`y_(1)=0.05 sin (3 pit-2x) and y_(2) = 0.05 sin (3pit+2x)`
Where x and y are in metres and t is in second. What
is the amplitude of the particle at `x = 0.5` m ? (Given,
`cos 57. 3 ^(@) = 0. 54)`
A. `2.7 ` cm
B. `5,4` cm
C. `8.1` cm
D. `10. 8` cm

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - a
Here, `y_(1)=0.05sin (3pit-2x), y_(2)=0.05 sin (3pit+2x)`
According to superposition principle, the resultant
displacements is
`y=y_(1)+y_(2) = 0.05 [sin (3pit-2x)+ sin (3pit+2x)`
or `y= 0.05 xx 2 sin 3pit cos 2x`
or, `y=(0.1 cos 2x) sin 3pit = R sin 3pit`
where,`R= 0.1 cos 2x=` amplitude of the resultant
standing wave.
At `x= 0.5` m
`R=0.1 cos 2x = 0.1 cos 2 xx 0.5`
`= 0.1 cos 1 ("radian") = 0.1 cos frac(180^(@))(pi) = 0.1 cos 57.3^(@)`
or `R=0.1xx054 m = 0.054 m = 5. 4 cm`