A string with linear mass density =0.0250kg/m under a tension of T=250. N is ori

Question

A string with linear mass density =0.0250kg/m under a tension of T=250. N is oriented in the x direction. Two transverse waves of equal amplitude and with a phase angle of zero(at t=0)but with different frequencies(=3000. rad/s and /3=1000. rad/s) are created in the string by an oscillator locator at x=0. The resulting waves, which travel in the positive x-direction, are reflected at a distant point, so there is a similar pair of waves traveling in the negative x-direction. Find the value of x at which the first two nodes in the standing wave are produced by these four waves.

Explanation / Answer

velocity of wave =t/u = (250/.025) = 100m/s

given 1 = 3000 rad/s

1 = v*2 / = 2/30 == > k1 = 2/ = 30

given 2 = 1000 rad/s

2 = v*2 / = 2/10 ==> k2 = 2/ = 10

hence wwe get sin wave travelling in positive direction = sin( kx-wt)

for first wave it is sin(30x-3000t )

for second wave it is sin(10x-1000t )

and similarly wave travellling in negatib=ve direction

for first wave it is sin(3000t + 30x)

for second wave it is sin(1000t +10x)

final wave caused due to addition of above four waves =

(sin(-3000t + 30x)+sin(3000t + 30x))+(sin(-1000t + 10x)+sin(1000t + 10x))

2sin(30x)cos(3000t)+2sin(10x)sin(1000x)

the first term 2sin(30x)cos(3000t) have nodes at 30x = n this gives x = 0,/30,2/30,3/30,4/30,5/30,6/30 ...

the second term 2sin(10x)cos(1000t) have nodes at 10x = n this gives x =0,the first term 2sin(30x)cos(3000t) have nodes at 30x = n this gives x = 0, /10,2/10,3/10 ...

the terms common in both are x = 0, /10,2/10,3/10 ...

hence the nodes occur at this point

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.