00:01
Hello friends.
00:02
In this question, we have a baby inside a mother's womb whose amplitude of heart beat is 1 .80 millimeters and the heart is beating with a frequency of 115 beats per minute.
00:14
Considering the heart walls beat with a simple harmonic motion, we need to find out the speed of the heart walls in the first part of the question.
00:25
But before we do that, let's make the units consistent.
00:28
That is, 1 .80 millimeters is equal.
00:31
Equals to 1 .80 times 10 to the power minus 2 meters.
00:38
And 115 bits per minute is 115 cycles per 60 seconds.
00:46
That is 115 divided by 60 hertz.
00:50
Now we can move to part a where we find the linear velocity, linear velocity of the simple harmonic motion that we write as s at, m will be using shm from now on so let's write it here this is simple harmonic motion vbt this is the velocity of the heartbeat and it changes with time this is equals to amplitude of beat multiplied with the angular velocity multiplied with sine of angular velocity and time red side this angular velocity of beat with a angular velocity that we denote by omega is equals to two pi times the frequency of the beat let's look at the sine of omega bt now sine theta goes from the bracket zero to one this is the period of the sine function so the maximum value for sign will be one and if you want to find the maximum value of the velocity what we do is b b t is equals to ab multiplied with omega b multiplied with 1 so this gives us we know the amplitude as 1 .80 times 10 to the power minus 3 meters and we multiply this with omega b that is the angular frequency which is 2 pi times 115 divided by 60 this is in hers.
02:49
On solving this, we get the maximum velocity of heart walls, that is vv, is equals to 2 .168 times 10 to the power minus 2 meters per second.
03:26
In part b, we are asked to find the frequency that is received by the heart.
03:32
We have a source from which the ultrasonic sound is being emitted so the frequency of the source which we call as f s is equals to two times 10 to the power 6 hertz and the velocity that the source has that we call at vs is equals to 0 because it's not moving at all and then we have the receiver that is the heart now this is the heart walls because is the heart walls on which the frequency is going.
04:13
We call it fr, the frequency received, and that is what we need to find.
04:18
We don't know it yet.
04:19
And here we are the velocity of the receiver.
04:22
We only consider the maximum velocity, and this is equals to the vb, that is the maximum velocity of the beat that we found earlier, which is 2 .168 times 10 to the power minus 2 meters per second...