For a particular transverse standing wave on a long string...

For a particular transverse standing wave on a long string, one of the antinodes is at $x=0$ and an adjacent node is at $x=0.10 \mathrm{m}$ . The displacement $y(t)$ of the string particle at $x=0$ is shown in Fig. $16-40$ where the scale of the $y$ axis is set by $y_{s}=4.0 \mathrm{cm} .$ When $t=0.50 \mathrm{s},$ what is the displacement of the string particle at $(\mathrm{a}) x=0.20 \mathrm{m}$ and ( b ) $x=0.30 \mathrm{m} ?$ What is the transverse velocity of the string particle at $x=0.20 \mathrm{m}$ at (c) $t=0.50$ s and $(\mathrm{d}) t=1.0 \mathrm{s} ?(\mathrm{e})$ Sketch the standing wave at $t=$ 0.50 s for the range $x=0$ to $x=0.40 \mathrm{m} .$

Transcript

00:02 Just to solve this problem find out the aperture is the a you can find out from the ys ys is equal the ys that is 4 centimeter so there is the atitude of that disresement then we have to find out the t the time and that has been taken to our time period is point five second that in that time that time total time will be in 0 .5 seconds that amplitude maximum is 4 cm so as we know four times of that time would be in the time period so four into this one is two second two second is the time period now let us go forward to the five if we assume the the way we will start from the initial point of view then that will be 0 degree.

01:23 I will take it in 0 to 5.

01:26 And then we have to find out the that is to be a lambda.

01:36 The lambda will be the web length.

01:39 The length will be 4.

01:40 The distance between the node and antimode.

01:43 That means that was the 1 .0 to 0 .1.

01:48 So it will be 0.

01:51 One if the lambda will be i guess the number over the part of the data another thing is that find out the omega that means the angular frequency if they twice five divided with two as you know the time period is two second and is two so the omega will be 3 .14 so radium second 3 .14 radiance second is omega.

02:37 Now we will get the total distance.

02:44 That is the displacement of the white is equal to a a sine omega t.

03:01 Plus 5 5 that means here we can get a that is the 4 centimeter sign and the omega will be 3 .1 t plus initial point we will take it if you want to find out the pie we can find it out from its initial x and y axis and just to get that the handle right there.

03:39 So the final this statement formula we will be the 4 sine 3 .14 t.

03:49 So that's going to be the final formula to find out the value of x and the value of t and the value of t when the value of t is 05 second that particular point that would be y t equal to sign 3 .1 0 .1 .5 seconds that will be the result but we want to find out and it's equal to point at x to point two so we have to put the value here that the writing will be 0 .2 sign that is 3 .1 into 0 .5 and also for x equal to 0 .5 and also for x equal to point three you can get that point point three because we know this x is related to the amplitude section and for that at that point we get the displacement total displacement y t so point three it is a sign sign 3 .14 into 5 5 5 will remain the same now let us find out the displacement that is the transverse velocity the transverse velocity will be vt equal to d y d y t of d that means with respect to time just differentiate with respect to time we get that that is a sign omega t will be after differentiation we will get the cause 3 .1 t we will get a value from this that will be like that 3 .14 the average it was given 4 3 .1d you will get the value and the appellate of this is 4 okay so that will be 3 .14 is the maximum amplitude.

07:40 We get this.

07:41 The final output will be like that...

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