![Solution!
B) To determine speed (rate of change) we take the derivative of the
displacement function with respect to time holding x as a constant
𝜕𝐷(𝑥,𝑡)
𝜕𝑡
=
7(𝑥−0.7𝑡)
[(𝑥−0.7𝑡)2+3]2
Now we can evaluate the speeds at (0.3m, 4.0s) and (1.4m, 4.0s)
7(0.3−0.7(4))
[(0.3−0.7(4))2+3]2 = -0.205m/s
7(1.4−0.7(4))
[(1.4−0.7(4))2+3]2 = -0.398m/s
Therefore a waves velocity can differ at any point depending on the
position and time.](https://image.slidesharecdn.com/learningobject1physics101-150209012812-conversion-gate02/85/Learning-object-1-physics-101-9-320.jpg)
Learning object 1 physics 101
- 1. Learning Object BY: ALYKHAN DEVSI
- 2. Transverse Speed and Wave speed This learning object will focus on the displacement function of a pulse and how we can use this to determine the velocity of the medium in which the pulse is travelling at a specific time and position. A continuously travelling wave can be thought of as a series of wave pulses.
- 3. Displacement of a pulse in One Dimension A waveform can be described by giving the displacement of each element of a medium at a fixed time. 𝐷 𝑥, 𝑡 = 2 (𝑥−𝑣𝑡)4+3
- 4. Warm up question! Which of the following equations describes a pulse moving towards the direction of decreasing x? A) 𝐷 𝑥, 𝑡 = 2 (𝑥−4𝑡)4+3 B) 𝐷 𝑥, 𝑡 = 2 (𝑥+9𝑡)4+3 C) 𝐷 𝑥, 𝑡 = 7 (𝑥−.6𝑡)4+3
- 5. Correct Answer is B! In the case of choice B, we see that in the denominator velocity multiplied by time is being added instead of subtracted from the position x of the pulse. 𝐷 𝑥, 𝑡 = 2 (𝑥−𝑣𝑡)4+3 vs. 𝐷 𝑥, 𝑡 = 2 (𝑥+𝑣𝑡)4+3 The presence of an addition symbol says that the pulse is moving towards the negative values of x, a good way to rationale this is by thinking about shifts of a graph.
- 6. Below are 3 graphs of the same function differing only by addition or subtraction symbols As you can see, the graph that has the addition on the denominator is shifted left, whereas that with the subtraction is shifted right (ie increasing x)
- 7. Lets try this question A pulse travelling through a string is described by : 𝐷 𝑥, 𝑡 = 5 (𝑥 − 0.7𝑡)2+3 A) What is the velocity of the pulse B) At t = 4.0s what are the speeds of the string elements located at x = 0.3m and x = 1.4m? (Hint: think about speed as the rate of change of displacement)
- 8. Solution! A) The equation for the displacement of a pulse is given by: 𝐷 𝑥, 𝑡 = 5 (𝑥−𝑣𝑡)2+3 The given equation was: 𝐷 𝑥, 𝑡 = 5 (𝑥−0.7𝑡)2+3 Therefore the velocity of this pulse is 0.7m/s
- 9. Solution! B) To determine speed (rate of change) we take the derivative of the displacement function with respect to time holding x as a constant 𝜕𝐷(𝑥,𝑡) 𝜕𝑡 = 7(𝑥−0.7𝑡) [(𝑥−0.7𝑡)2+3]2 Now we can evaluate the speeds at (0.3m, 4.0s) and (1.4m, 4.0s) 7(0.3−0.7(4)) [(0.3−0.7(4))2+3]2 = -0.205m/s 7(1.4−0.7(4)) [(1.4−0.7(4))2+3]2 = -0.398m/s Therefore a waves velocity can differ at any point depending on the position and time.
- 10. Thank You!