Calculating Displacement, Velocity, and Acceleration of a Harmonic Wave Train

AI Thread Summary
A simple harmonic wave train with an amplitude of 3 cm, frequency of 200 Hz, and a velocity of 20 m/s is analyzed to find the displacement of a particle at 50 cm from the origin after 2 seconds. The calculated displacement using the equation y(x, t) = Acos(2πf(x/v - t)) yields 0.0236 m, while the textbook states the answer is 0.02523 m. Confusion arises regarding the use of radians versus degrees for the cosine function, with the consensus that radians should be used for accurate results. The discussion highlights the importance of evaluating large angles carefully and suggests that the problem may lack clarity regarding the wave's phase shift. Ultimately, the correct approach to the calculation is emphasized, focusing on radians and precision in evaluating trigonometric functions.
Nishikino Maki
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Homework Statement


A simple harmonic wave train of amplitude 3 cm and frequency 200 Hz travels in the +ve direction of x-axis with a velocity of 20 m/s. Calculate the displacement, velocity, and acceleration of a particle situated at 50 cm from the origin at t = 2 s.

Homework Equations


I used y(x, t) = Acos(2\pi f(\frac{x}{v}-t))

The Attempt at a Solution


Plugging in the values into the above equation, I got y(0.5, 2) = 0.03cos(400\pi (\frac{.5}{20} - 2)), which evaluates to 0.0236 m. However, the book says the answer is 0.02523 m.

This is marked as an easy question and is one of the first ones, so I think that I'm missing something basic?
 
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You need to be a bit careful evaluating trig functions of large angles. The series expansions used by calculators get rather inaccurate. Instead, first reduce the angle to something less than 2 pi. I think you'll find both your answer and the given answer rather inaccurate!
 
Part of my confusion lies in whether I should be in radians or degrees - the examples in the book all use radians, but when I evaluated the above, I got cos(-790\pi), or 1. Changing the angle to something less than 2\pi still gets me 1.

When I use degrees, I get 0.0236, which is closer to 0.02523. I plugged the expression into Wolfram-Alpha, which got me the same as the one on my calculator. Changing the angle to something less than 360 degrees got me 0.0271.
 
Nishikino Maki said:
Part of my confusion lies in whether I should be in radians or degrees - the examples in the book all use radians, but when I evaluated the above, I got cos(-790\pi), or 1.
Definitely radians. The standard equation you quoted assumes radians.
It gives you 1 for the value of the cos function, but you still have to multiply by A.
 
Nishikino Maki said:

Homework Statement


A simple harmonic wave train of amplitude 3 cm and frequency 200 Hz travels in the +ve direction of x-axis with a velocity of 20 m/s. Calculate the displacement, velocity, and acceleration of a particle situated at 50 cm from the origin at t = 2 s.

Homework Equations


I used y(x, t) = Acos(2\pi f(\frac{x}{v}-t))

The Attempt at a Solution


Plugging in the values into the above equation, I got y(0.5, 2) = 0.03cos(400\pi (\frac{.5}{20} - 2)), which evaluates to 0.0236 m. However, the book says the answer is 0.02523 m.

This is marked as an easy question and is one of the first ones, so I think that I'm missing something basic?
I got the same answer as you.
BTW the problem should state that the wave is inded ~ cos(kx - wt) and not something like cos(kx - wt + φ), φ ≠ 0.
Part of my confusion lies in whether I should be in radians or degrees
Always assume radians. And always assume natural instead of base-10 logs. Calculus falls apart otherwise!
 
rude man said:
I got the same answer as you.
Then you must be using a calculator that truncates the precision of pi at the same point. The right answer is clearly 0.03m.
 
haruspex said:
Then you must be using a calculator that truncates the precision of pi at the same point. The right answer is clearly 0.03m.
Not truncate. Round off.
But yes, score one for the Aussies. Again, only if the wave is cos(kx - wt + φ), φ = 0 assumed. The problem is not clearly stated.
 
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