Skip to main content
Ch 15: Mechanical Waves

Chapter 15, Problem 15.34b

Two pulses are moving in opposite directions at 1.0 cm/s on a taut string, as shown in Fig. E15.34. Each square is 1.0 cm. <IMAGE> Sketch the shape of the string at the end of (b) 7.0 s.

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
97
views
Was this helpful?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first stop, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A top rope as shown below has two pulses moving towards each other at a speed of 2.0 centimeters per second sketch the shape of the rope at the end of 3.0 seconds. So that's our angle. Our angle is we're trying to figure out how to draw a sketch of our rope to denote what will happen. Once these two pulses that are moving towards each other, what's gonna happen after the end of three seconds passing. So with that in mind, our first step is is we need to note where the initial positions of the pulses are. So to keep things simple looking at our graph that's provided to us by the pro up, we have our two pulses, we have our pulse on the left that is moving to the right at 2.0 centimeters per second. And we have our pulse on the right that is moving to the left towards our left pulse at 2.0 centimeters per second. And these both of these speeds are denoted by orange arrows pointing towards each other. And as we could see looking at our left pulse, it's centered at eight centimeters and we have our right pulse, which is that centered at 20 centimeters. Awesome. So now at this point, we need to consider the movement of the pulses. So let us note that each pulse is moving at a speed of 2.0 centimeters per second, which we can recall and write this equation at the distance D is equal to the speed multiplied by the time. So with that in mind, we can now investigate the shape of the rope after 3.0 seconds have passed. So let us note that in 3.0 seconds, each pulse will move to the following new positions, which I'm gonna mark these new positions in green. So focusing on our left pulse first, it will move six centimeters to the right. So starting at six, that's where the beginning of our upside down shaped parabola begin. So you go from six and we go seven is 18, is 29, is three, 10 is so it goes from six. So 1234567. So from six. So starting from six, we need to skip one. So it goes 123456. So it starts at 12 and then it will end up centered at 14. So that means skip one, that means the other end of her bump will be at 16 and the right pulse will move six centimeters to the right. So, so looking at our 18 here, so we go 123456. So step at 12 and it'll be end up centered at 14 centimeters. So as you can see, as you were probably noticing, therefore, since the pulses are moving towards each other, they will eventually meet and overlap completely through constructive interference. As you could see when we were mapping out our skeleton points. So also we need to note that both pulses will be centered at 14 centimeters and the combined height will be the sum of the individual pulses. So as you could see with our little grid here it goes 12 and 312 and three. So each bump has a height of three. So three plus three is six. So that means our final bump, our final graph will have a height of six for the combined height. Thus, we can do our final sketch of our graph like. So, so looking at our bumps that we were comparing up above down below on our empty graph, so we know that one end of our pulse is at 12 and the other end is at six. So we know it's centered at 14. So we need to go up six. So 12345. And six. So at the very top of this rack, and then we need to draw lines to denote a rope and green connecting R two sides of her pulse here. And then we need to draw a Parabola shape or upside down parabola shape to denote our rope. So it's a little bit bumpy here. But that is literally what our final sketch should look like because both of our pulses will combine to form one pulse and looking at a more professional, more computer generated like graph, our final graph should look something like this. So that means we did our final sketch graph correctly and that's it. That's how you solve for this problem. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.
Related Practice
Textbook Question
A wire with mass 40.0 g is stretched so that its ends are tied down at points 80.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude at the antinodes of 0.300 cm. (a) What is the speed of propagation of transverse waves in the wire?
584
views
Textbook Question
A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. (a) What is the frequency of its fundamental mode of vibration?
704
views
Textbook Question
One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the break-ing stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?
1477
views
Textbook Question
Two loudspeakers, A and B (Fig. E16.35)

, are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. What is the lowest frequency for which (a) constructive interference occurs at point Q

1125
views
Textbook Question
Two loudspeakers, A and B (Fig. E16.35)

, are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. What is the lowest frequency for which (b) destructive interference occurs at point Q?

506
views
Textbook Question
Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone arranged as shown in

Fig. E16.39. For what frequencies does their sound at the speakers produce (a) constructive interference

802
views