Suppose two waves, identical except for the direction of travel, approach each o

Question

Suppose two waves, identical except for the direction of travel, approach each other in a medium that obeys Hooke's law. At certain instants, the waves interfere constructively at all positions. Suppose that the kinetic energy in each of the initial waves is K.

At one such instant, when their peaks align, determine the kinetic energy of the wave function that results from the superposition of both waves.

Express your answer in terms of K.

Suppose two waves, identical except for the direction of travel, approach each other in a medium that obeys Hooke's law. At certain instants the waves interfere constructively at all positions. Suppose that the kinetic energy in each of the initial waves is K. At one such instant, when their peaks align, determine the kinetic energy of the wave function that results from the superposition of both waves.

Explanation / Answer

Let us consider the two waves are on a string. Let us take a small particle of mass m of the string into consideration.

Now let the displacement of the particle due to the two waves be y1 and y2.

Now, when the two waves interfere constructively the resultant displacement y of the particle is:

y = y1 + y2 (from principle of superposition)

or, dy/dt = dy1/dt + dy2/dt

or v = v1 + v2

where v = resultant velocity of the particle after interference and v1 and v2 are the individual velocities due to the two waves.

Now for the particle, its kinetic energy KE due to individual waves when other is not present is

KE1 = (1/2)m(v1)2 = KE of the particle due to first wave only, similarly

KE2 = (1/2)m(v2)2 = KE of the particle due to second wave only

Now it's given that KE1 = KE2 = K

So we have (1/2)m(v1)2 = (1/2)m(v2)2

or (v1)2 = v2)2

or lv1l = lv2l , thus both speeds are equal.

Now when the particle experiences both waves simultaneously during the constructive experience we saw that its velocity becomes

v = v1 + v2

but since the waves are identical except for the direction of travel, we have velocities v1 = v2

So the new KE of the particle = (1/2)mv2 = (1/2)m(v1 + v2)2 = (1/2)m (2v1)2 (v1 = v2)

or the new KE of the particle = (1/2)m (2v1)2 = 4.[(1/2)m (v1)2] = 4K

This can be said about all such small particles on the string and thus abou the whole wave.

So, the kinetic energy of the wave function that results from the superposition of both waves = 4K.

This concludes the answers. If you find anything lacking please let me know.. I will resolve your query without delay...

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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