Question

Two wave pulses travel on a string toward each other. The wave pulses can be described as y1 = 5/(((kx − ωt)^2) 2) and y2 = −5/(((kx + ωt − 6)2 )+2)' , where k = 1 rad/m and ω = 8 rad/s. At what instant do the two cancel everywhere? (Assume x is in meters and t is in seconds.)

Answer 1

To determine the instant when two wave pulses with functions y₁ and y₂ cancel each other everywhere, set the inside of the square brackets in the denominators equal to each other and solve for t, which yields t = 3/8 seconds.

Step-by-step explanation:

Given the wave functions y₁ = 5/(((kx − ωt)²) + 2) and y₂ = − 5/(((kx + ωt − 6)²) + 2), with k = 1 rad/m and ω = 8 rad/s, we need to find the instant when these two wave pulses cancel each other everywhere.

To find when the waves cancel each other out, we set the functions equal to each other and solve for t. In this case, the waves have inversely proportional amplitudes, so they will cancel out when the value inside the brackets of the denominator is the same, meaning that the brackets themselves must be equal as their magnitudes are multiplicative inverses due to the negative sign in y₂.

Setting the expressions in the brackets equal to each other, we have:

(kx − ωt)² = (kx + ωt − 6)²

Substituting the given values for k and ω and simplifying, we'll find that t = 3/8 seconds is the time when the two pulses cancel everywhere.