Question

A standing wave on a string is given by: y(x,t)=0.04sin⁡(2πx)cos⁡(100πt), where x and y are in meters and t is in seconds. The left end of the string is located at x=0. The location of the 5th antinode is at

Answer 1

The location of the 5th antinode in the given standing wave y(x, t) = 0.04 sin(2πx) cos(100πt) is at x = 2.25 meters.

Step-by-step explanation:

The question pertains to the location of the 5th antinode in a standing wave on a string. The equation given for the wave is y(x, t) = 0.04 sin(2πx) cos(100πt), where the sin function argument indicates the spatial component and the cos function indicates the temporal component of the wave.

In standing waves, antinodes occur at positions where the sine function equals 1 or -1. This happens at odd multiples of π/2 within the sine function's argument. The fifth antinode would occur at (2n - 1)(π/2) where n is the antinode number, so for the 5th antinode, n=5. Substituting into the equation, we have 2πx = (2*5 - 1)(π/2), therefore x = (9π/2) / (2π) = 9/4.

Thus, the location of the 5th antinode is at x = 9/4 meters or 2.25 meters.