00:01
For this problem on the topic of wave motion, we are given the wave function for a wave on a string, and we want to use this wave function to find the transfer speed and acceleration of the string at 0 .2 seconds for a point on the string that is located at 1 .6 meters.
00:17
Next, we want to find the wavelength, period, and speed of propagation for this wave.
00:23
Now, the wave function is given as y is equal to 0 .12 meters times the sine of pi over 8 times x, plus 4 pi times t.
00:33
So the transfer speed for a point on this wave, v is equal to d .y d t.
00:41
And this is equal to the time derivative of the function y.
00:49
And so this is 0 .12 times 4 pi times the cosine of pi by 8 times x plus 4 pi times t.
01:07
And so if we were to calculate the velocity at t equal to 0 .2 seconds and x equal to 1 .6 meters, we get the speed of a particle at this point by substituting t is equal to 0 .2 and x 1 .6 into the above equation to be minus 1 .51 meters per second.
01:34
So that's the speed of a particle at 1 .6 meters after 0 .2 seconds...