Test: Waves and Sound - 2 - JEE MCQ

The resultant displacement is given by y = y₁ + y₂ = 0.05{sin(3πt − 2x) + sin(3πt + 2x)}
Using the trigonometric relation sin(α + β) + sin(α − β) = 2sin αcos β
we have y = 0.1cos2x⋅sin3πt
or y = Rsin 3πt
where R, the amplitude of standing waves, is given by
R = 0.1cos2x
When x = 0.5 m
cos 2x = cos(2 × 0.5rad)
= cos(1rad)
= cos57.3^∘ = 0.54
∴ Amplitude R at x = 0.5 is 0.1 × 0.54 = 0.054 m = 5.4 cm

y = y₁ + y₂
= 5[sin(3πt − 2πx) + sin(3πt + 2πx)]
Using sin(α + β) + sin(α − β) = 2 sin α cos β, we get
y = 10cos(2πx) sin(3πt)
⇒ y = A sin(3πt), where A = 10 cos(2πx)
Amplitude A at x = 2 cm is
10 × cos(2π × 2) = 10 cos 4π = 10 cm

We know

But specific gravity 8 
⇒ 
∴  
But  
∴  

Fundamental frequency 

v.t

Intensity∝(amplitude)²
∝(2acoskx)²
Hence, intensity will be maximum when cos kx  is maximum.

Let the string vibrates in p loops, wavelength of the mode of vibration is given by
Given,
or
Comparing it with standard equation, we get and