00:01
Now in this question it is given, the equation of the transverse wave, that is yx t, is equals to 15 .0, sine pi x divided by 8 minus 4 by t.
00:12
Now for question number a, we have the value of x to be 6 centimeter or 0 .06 meter, and the value of time is 0 .250 seconds.
00:24
Now, we know the velocity of transverse wave that is u is equals to dy by dt, that is minus 4 pi times rather than 15 we will be writing it in meter, that is 0 .15 meter, cos pi x divided by 8 minus 4 pi t.
00:46
Putting in the values we'll have, u is equals to minus 4 pi times 0 .15, cost pi times 0 .15, cost pi times 0 .5.
00:55
0 .06 divided by 8 minus 4 pi 0 .25 seconds.
01:01
This will give us the value of transverse velocity to be u equals to 1 .88 meter per second.
01:10
Now for question number b we know that the general form of for the transfer speed is given by u equals to dy by dt equals to minus omega y m cost k x plus minus omega t.
01:26
Now for maximum value of u, that is u m, will be equals to omega ym.
01:38
Now putting in the value of omega from the question, that is 4 pi radian per second, times the value of ym, which is 15 centimeter, which will turn into 0 .15 meter...