Question

two simple harmonic progressive waves are represented by Y₁ = 2 sin 2π (100t - x/60) cm and Y₂ = 2 sin 2π (100t + x/60) cm. The waves combine to form a stationary wave. Find (i) amplitude at antinode (ii) distance between adjacent node and adjacent antinode. (iii) loop length (iv) wave velocity

Answer 1

Given :

y1 = 2 sin 2π[100t –x/ 60 ]cm

y2= y1 = 2 sin 2π[ 100t +x/ 60] cm

Formula : Y= R sin2 πnt

Where R=2A Cos [2 πx/ λ]

Now resultant equation is  

Y= y1+y2

=2sin2 π (100t-x/60] + 2sin 2 π[100t+x/60]

Y=2x2sin2 π(100t)cos2 π(x/60]

Y=4cos (2 πx/60)sin 2 π(100t)

On comparing the above equation with  

Y=Rsin(2 πt)

4cos (2πx/60)=R

But R=2 πcos(2 πR/ λ)

Since λ=60cm

R is maximum when Cos (2 πR/ λ)=1

R=4x1=4cm

Λ=60m

λ/4=60/4=15cm

Therefore distance between  successive node and anti node is 15cm

Length= λ/2=60/2=30cm

Similarly =v=n λ

N=100hz

A=60cm

V=100x60=600cm/s

V=60m/s

Answer 2

Given :

y1 = 2 sin 2π[100t –x/ 60 ]cm

y2= y1 = 2 sin 2π[ 100t +x/ 60] cm

Formula : Y= R sin2 πnt

Where R=2A Cos [2 πx/ λ]

Now resultant equation is  

Y= y1+y2

=2sin2 π (100t-x/60] + 2sin 2 π[100t+x/60]

Y=2x2sin2 π(100t)cos2 π(x/60]

Y=4cos (2 πx/60)sin 2 π(100t)

On comparing the above equation with  

Y=Rsin(2 πt)

4cos (2πx/60)=R

But R=2 πcos(2 πR/ λ)

Since λ=60cm

R is maximum when Cos (2 πR/ λ)=1

R=4x1=4cm

Λ=60m

λ/4=60/4=15cm

Therefore distance between  successive node and anti node is 15cm

Length= λ/2=60/2=30cm

Similarly =v=n λ

N=100hz

A=60cm

V=100x60=600cm/s

V=60m/s
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