two sources S2 and S2 , separated by 2 m , according to equation y1 = 0.03 sin( pi*t) and y2 = 0.02 sin(pi*t) , where y1and y2 and t are in S.I units . they send out waves of equal velocity 1.5 m/s . Calcu;ate the amplitude of the resultant motion of the particle co- linear with S1 and S2 and located at a point
1) P1 to the right of S1
2) P2 to the left of S2
3) P3 in the middle of S1and S2
Answers
Answer:
1.
Here,v=1.5ms−1Let the position of P1 be at x distance away from S1So distance of P1 from S2= x−2Time taken by wave to reach P1 from S1 =t't'=xv=x1.5Time taken by wave to reach P2 from S2 =t''t''=x−2v=x−21.5So the wave eqn. at P1 due to S1 and S2 will bey1=0.03sin(π(t−t'))y2=0.02sin(π(t−t'')) i. Resultant wave at P1y=y1+y2=0.03sin(π(t−t'))+0.02sin(π(t−t''))=0.03sin(π(t−x1.5))+0.02sin(π(t−x−21.5))=0.03sin(πt−2πx3)+0.02sin((πt−2πx3)+4π3)=0.03sin(πt−2πx3)+0.02sin(πt−2πx3)cos(4π3)+0.02cos(πt−2πx3)sin(4π3)sin(4π3)=−3√2, cos(4π3)=−0.5=(0.03-0.02×0.5)sin(πt−2πx3)−0.01×3–√cos(πt−2πx3)The eqn. becomes y=Rsin(πt−2πx3−ϕ)−−−1.Let Rcosϕ=0.02Rsinϕ=−3√100tanϕ=−3√2Thus, from the theory of interference we have amplitude of the wave at P1=RR=0.022+(−3√100)2−−−−−−−−−−−−−√=7√100=0.026Here,v=1.5ms−1Let the position of P1 be at x distance away from S1So distance of P1 from S2= x-2Time taken by wave to reach P1 from S1 =t't'=xv=x1.5Time taken by wave to reach P2 from S2 =t''t''=x−2v=x−21.5So the wave eqn. at P1 due to S1 and S2 will bey1=0.03sin(π(t−t'))y2=0.02sin(π(t−t'')) i. Resultant wave at P1y=y1+y2=0.03sin(π(t−t'))+0.02sin(π(t−t''))=0.03sin(π(t−x1.5))+0.02sin(π(t−x−21.5))=0.03sin(πt−2πx3)+0.02sin((πt−2πx3)+4π3)=0.03sin(πt−2πx3)+0.02sin(πt−2πx3)cos(4π3)+0.02cos(πt−2πx3)sin(4π3)sin(4π3)=−32, cos(4π3)=−0.5=(0.03-0.02×0.5)sin(πt−2πx3)−0.01×3cos(πt−2πx3)The eqn. becomes y=Rsin(πt−2πx3−ϕ)−−−1.Let Rcosϕ=0.02Rsinϕ=−3100tanϕ=−32Thus, from the theory of interference we have amplitude of the wave at P1=RR=0.022+(−3100)2=7100=0.026
2.
Here,v=1.5ms−1Let the position of P1 be at x distance away from S1So distance of P2 from S2= x+2Time taken by wave to reach P2 from S1 =t't'=xv=x1.5Time taken by wave to reach P2 from S2 =t''t''=x+2v=x+21.5So the wave eqn. at P2 due to S1 and S2 will bey1=0.03sin(π(t−t'))y2=0.02sin(π(t−t'')) i. Resultant wave at P1y=y1+y2=0.03sin(π(t−t'))+0.02sin(π(t−t''))=0.03sin(π(t−x1.5))+0.02sin(π(t−x+21.5))=0.03sin(πt−2πx3)+0.02sin((πt−2πx3)−4π3)=0.03sin(πt−2πx3)+0.02sin(πt−2πx3)cos(4π3)−0.02cos(πt−2πx3)sin(4π3)sin(4π3)=−3√2, cos(4π3)=−0.5=(0.03-0.02×0.5)sin(πt−2πx3)+0.01×3–√cos(πt−2πx3)Let Rcosϕ=0.02Rsinϕ=3√100The eqn. becomes y=Rsin(πt−2πx3+ϕ)−−−2.Thus, from the theory of interference we have amplitude of the wave at P2=RR=0.022+(3√100)2−−−−−−−−−−−−√=7√100=0.026mComparing eqn. 1 and 2. we see that the two waves are out of phase ϕSo, in 1st we have -0.026 mand in the second we have 0.026 m Here,v=1.5ms−1Let the position of P1 be at x distance away from S1So distance of P2 from S2= x+2Time taken by wave to reach P2 from S1 =t't'=xv=x1.5Time taken by wave to reach P2 from S2 =t''t''=x+2v=x+21.5So the wave eqn. at P2 due to S1 and S2 will bey1=0.03sin(π(t−t'))y2=0.02sin(π(t−t'')) i. Resultant wave at P1y=y1+y2=0.03sin(π(t−t'))+0.02sin(π(t−t''))=0.03sin(π(t−x1.5))+0.02sin(π(t−x+21.5))=0.03sin(πt−2πx3)+0.02sin((πt−2πx3)−4π3)=0.03sin(πt−2πx3)+0.02sin(πt−2πx3)cos(4π3)−0.02cos(πt−2πx3)sin(4π3)sin(4π3)=−32, cos(4π3)=−0.5=(0.03-0.02×0.5)sin(πt−2πx3)+0.01×3cos(πt−2πx3)Let Rcosϕ=0.02Rsinϕ=3100The eqn. becomes y=Rsin(πt−2πx3+ϕ)−−−2.Thus, from the theory of interference we have amplitude of the wave at P2=RR=0.022+(3100)2=7100=0.026mComparing eqn. 1 and 2. we see that the two waves are out of phase ϕSo, in 1st we have -0.026 mand in the second we have 0.026 m
3. Try to solve the third in the same way as above.