Two waves in a string are defined by the wave functions: y1(x, t) 2.0sin (20.0x-

Question

Two waves in a string are defined by the wave functions:
y1(x, t) 2.0sin (20.0x-32.0t)

and
y2(x,t)=2.0sin(25.0x-40.0t)
where x, y1, and y2 are measured in centimeters and t in seconds. (a) What is the phase difference between these two waves at the point x = 5.0 cm at the instant in time t = 2.0 s? (b) Find a location where the phase difference between the two waves is equal to +pi at t = 2.0 s. Two waves in a string are defined by the wave functions:
y1(x, t) 2.0sin (20.0x-32.0t)

and
y2(x,t)=2.0sin(25.0x-40.0t)
where x, y1, and y2 are measured in centimeters and t in seconds. (a) What is the phase difference between these two waves at the point x = 5.0 cm at the instant in time t = 2.0 s? (b) Find a location where the phase difference between the two waves is equal to +pi at t = 2.0 s. Two waves in a string are defined by the wave functions:
y1(x, t) 2.0sin (20.0x-32.0t)

and
y2(x,t)=2.0sin(25.0x-40.0t)
where x, y1, and y2 are measured in centimeters and t in seconds. (a) What is the phase difference between these two waves at the point x = 5.0 cm at the instant in time t = 2.0 s? (b) Find a location where the phase difference between the two waves is equal to +pi at t = 2.0 s.

Explanation / Answer

Acccording to the given problem,

(a)
Phase difference
= l(20x - 32t) - (25x - 40t)l
Plugging, x = 5 cm and t = 2 sec,
= l(100 - 64) - (125 - 80)l
= 9 radian
= 516°
Since the phase difference has to be < 360°
answer is 516 - 360 = 156°.

(b)
Phase of the first wave after 2 sec = 20x - 64 and
phase of the second wave after 2 sec = 25x - 80
Their difference = 5x - 16 = (2k ± 1)
=> 5x = 16 + (2k ± 1) = (5.093 + 2k ± 1)
Least value of 5x is when k = - 2
=> 5x = 0.093
=> x = (0.093)/5 = 0.0584 cm.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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