00:01
Hi, in this question the wave function of the wave, y of x comma t, is given as 0 .280 sine 8 pi t minus 3 pi x plus pi by 4.
00:27
So let's call this equation number 1.
00:31
Now let's consider the standard form of the wave function.
00:39
Which is y of x comma t is equal to a sine k x minus omega t plus phi and let's call this equation number two so now we'll just compare equation number one and two so compare one and two so upon comparing their coefficients we can rate that a that is the amplitude is 0 .280 meters.
01:19
The value of k is minus 3 pi newton per meter is the units.
01:26
Space difference is pi by 4 and the value of omega is minus 8 pi.
01:36
Now we'll move on to the first sub -part of this question in which we need to find out the average rate at which energy is transmitted along the string.
01:54
So this is given by half mu omega square a square into v.
02:04
Now v is nothing but omega by k.
02:11
So we know the values for omega and k.
02:14
So this will become minus 8 pi divided by minus 3 .5.
02:20
Which will give 2 .67 meter per second.
02:27
So this is the velocity of the wave.
02:31
And from the question, the linear mass density mu has a value of 84 .5 gram per meter, which is equal to 84 .5 into 10 to the power minus 3 kilogram per meter.
02:51
So now upon substituting the values, the average rate will become half into mu is 84 .5 into 10 to the power minus 3 into omega is minus 8 5 whole square into amplitude of 0 .280 whole square into velocity v which is 2 .67...