A sinusoidal wave is traveling on a string with speed 30.00 cm/s. The displacement of the particles of the string at x = 15 cm is found to vary with time according to the equation y = (5.0 cm) sin[1.5 - (3.0 s-1)t]. The linear density of the string is 5.0 g/cm.
(a) What is the frequency of the wave?
(b) What is the wavelength of the wave?
(c) Give the general equation giving the transverse displacement of the particles of the string as a function of position and time.
(d) Calculate the tension in the string.
Unless you have a typo, sin[1.5 - (3.0 s-1)t] is really sin[1.5 - 2t]
when is 2 t equal to 2 pi ?? (that is when the argument of the sin function changes by 2 pi, a period)
That is when t increases by pi
so the period, T is pi
the frequency = 1/pi
wavelength is how far it goes in pi seconds
That is 30 * pi
y = (5.0 cm) sin[1.5 - 2t] sin (2 pi x/L - phi)
where L is that wavelength
get the phase angle phi from the value of y when x = 15 and t = 0
you have the speed of the wave c = L/T
You know the equation for speed of the wave versus tension and density, so calculate the tension.
To find the answers to these questions, we'll use some key formulas and concepts. Let's break it down step by step:
(a) The frequency (f) of a wave is the number of oscillations per unit time. In this case, the equation for the wave is given as y = (5.0 cm) sin[1.5 - (3.0 s^-1)t].
The argument inside the sin function represents the angular frequency (ω) of the wave. In this case, ω = 3.0 s^-1.
The relationship between angular frequency (ω) and frequency (f) is given by the equation f = ω / (2π). Plugging in the value of ω, we get:
f = 3.0 s^-1 / (2π) ≈ 0.477 s^-1
So, the frequency of the wave is approximately 0.477 Hz.
(b) The speed (v) of a wave is related to its wavelength (λ) and frequency (f) by the equation v = λf.
Given that the speed of the wave is 30.00 cm/s and the frequency is approximately 0.477 Hz, we can rearrange the equation to solve for the wavelength:
λ = v / f = 30.00 cm/s / 0.477 s^-1 ≈ 62.96 cm
So, the wavelength of the wave is approximately 62.96 cm.
(c) The general equation giving the transverse displacement of the particles of the string as a function of position (x) and time (t) for a sinusoidal wave can be expressed as:
y(x, t) = A sin(kx - ωt + φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency, and φ is the phase constant.
In our given equation y = (5.0 cm) sin[1.5 - (3.0 s^-1)t], we can determine that the amplitude is 5.0 cm. The wave number (k) can be calculated using the relationship k = 2π / λ, where λ is the wavelength we found in part (b).
Calculating k:
k = 2π / λ = 2π / 62.96 cm ≈ 0.100 cm^-1
Now we have all the required values to express the general equation:
y(x, t) = (5.0 cm) sin(0.100 cm^-1 * x - (3.0 s^-1) * t + φ)
(d) To calculate the tension in the string, we can use the formula:
Tension (T) = (Linear density (μ) * wave speed (v)^2)
Given that the linear density of the string is 5.0 g/cm and the wave speed is 30.00 cm/s, we have:
T = (5.0 g/cm) * (30.00 cm/s)^2
Converting grams to kilograms:
T = (5.0 g/cm) * (30.00 cm/s)^2 = (5.0 * 10^-3 kg/cm) * (30.00 cm/s)^2
Simplifying the units:
T = (5.0 * 10^-3 kg/cm) * (30.00 cm/s)^2 = (5.0 * 10^-3 kg/cm) * (900.00 cm^2/s^2)
T = 4.50 N
Therefore, the tension in the string is 4.50 N.