A harmonic wave with a frequency of 90 Hz and an amplitude of 0.025 m travels along a string to the right with a speed of 20 m/s.

a)Write a suitable sine wave function for this wave in terms of x and t.
b) Find the maximum speed of a point on the string.
(c) Find the maximum acceleration of a point on the string.

y = .025 sin (2 pi f t - 2 pi x/L)

L = v T = v/f
so
y = .025 sin 2 pi f t - 2 pi x f/v)
= .025 sin 2 pi f( t - x/v)
= .025 sin 2 pi *90 (t - x/20)
= .025 sin 2 pi (90 t - 4.5 x)
-------------------------------------
dy/dt = speed = 180 pi *.025 cos 2 pi (90 t - 4.5 x)
d^2y/dt^2 = acceleration =- .025 (180 pi)^2 sin 2 pi (90 t - 4.5 x) = -(180 pi)^2 y

a) To write a suitable sine wave function for this wave in terms of x and t, we can use the general equation for a harmonic wave:

y(x, t) = A * sin(kx - ωt + φ)

Where:
- y(x, t) is the displacement of the wave at a point x and time t,
- A is the amplitude of the wave,
- k is the wave number (k = 2π / λ, where λ is the wavelength),
- x is the position along the string,
- ω is the angular frequency (ω = 2πf, where f is the frequency),
- t is the time, and
- φ is the phase constant.

Given:
- Frequency (f) = 90 Hz
- Amplitude (A) = 0.025 m
- Wave speed (v) = 20 m/s

We can start by finding the wavelength (λ) using the wave speed and frequency:

v = λf
λ = v / f
λ = 20 m/s / 90 Hz

Next, we can find the wave number (k) using the equation k = 2π / λ:

k = 2π / λ
k = 2π / (20 m/s / 90 Hz)

Once we have the wave number, we can write the sine wave function:

y(x, t) = A * sin(kx - ωt + φ)

b) To find the maximum speed of a point on the string, we can take the derivative of y(x, t) with respect to time (t), and then find the amplitude of this derivative.

Since we already have the sine wave function:

y(x, t) = A * sin(kx - ωt + φ)

We can differentiate it with respect to time:

dy(x, t) / dt = -ωA * cos(kx - ωt + φ)

The maximum speed occurs when the cosine term takes on its maximum value of 1. So, the maximum speed (v_max) is given by:

v_max = ωA

We can find ω using the relationship ω = 2πf, where f is the frequency. So:

ω = 2πf
ω = 2π * 90 Hz

Finally, we can substitute the values of ω and A to find the maximum speed (v_max).

c) To find the maximum acceleration of a point on the string, we can take the derivative of the velocity (which we found in part b) with respect to time (t), and then find the amplitude of this derivative.

So, we differentiate the velocity (v) with respect to time:

dv(x, t) / dt = d(ωA) / dt
dv(x, t) / dt = 0 (since ω and A are constants)

The maximum acceleration occurs when the derivative is at a maximum, which in this case, is 0. Therefore, the maximum acceleration (a_max) is 0.

To summarize:
a) The suitable sine wave function for this wave in terms of x and t is: y(x, t) = 0.025 * sin(kx - ωt + φ).
b) The maximum speed of a point on the string is v_max = ωA.
c) The maximum acceleration of a point on the string is a_max = 0.