Calculating Displacement and Speed in a Standing Wave on a String

[Lakers08]

Adjacent antinodes of a standing wave on a string are a distance 14.5 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 8.00×10-2 s . The string lies along the + x - axis and is fixed at x = 0.

Find the displacement of a point on the string as a function of position and time.
Find the speed of propagation of a transverse wave in the string.
Find the amplitude at a point a distance 3.20 to the right of an antinode.


- the first question was pretty easy easy i just used the formula :
A*cos(2*pi*t/T)*sin(2*pi*x/(2*s)) T = period A = amplitude
and i got the result: 3.25*10^(-3)m. which is correct, now can anybody give me any clues on how to calculate the speed of the propagation and the amplitude at 3.20 to the right of the antinode?

 

[Pyrrhus]

if i remember correctly

v = \lambda f

v = \frac{w}{k}

For the third question you know an atinode is located at a distance of \frac{\lambda}{4} so x = \frac{\lambda}{4} + 3.2

Also
k = \frac{2 \pi}{\lambda}

Remember Stationary waves formula

y = (2A \sin kx) \cos \omega t

 

[wais]

To calculate the speed of propagation, we can use the formula v = λf, where v is the speed, λ is the wavelength, and f is the frequency. In this case, the wavelength is equal to the distance between adjacent antinodes, which is 14.5 cm (0.145 m). The frequency can be calculated using the formula f = 1/T, where T is the period given in the problem. Plugging in the values, we get f = 1/8.00×10-2 s = 12.5 Hz. Now, we can calculate the speed using v = 0.145 m * 12.5 Hz = 1.81 m/s.

To find the amplitude at a point 3.20 cm to the right of the antinode, we can use the formula A*cos(2*pi*t/T)*sin(2*pi*x/(2*s)). In this case, the value of x is 0.032 m (3.20 cm converted to meters) and all the other values remain the same. Plugging in these values, we get A*cos(2*pi*t/T)*sin(2*pi*0.032/(2*0.145)) = 0.850*cos(2*pi*t/8.00×10-2)*sin(2*pi*0.032/0.290) = 0.850*cos(25*pi*t)*sin(0.219*pi). This is the amplitude at a point 3.20 cm to the right of the antinode.