. A long string carries a wave; a 6-m segment of the string contains four complete wavelengths and has a mass of 180 g. The string vibrates sinusoidal with a frequency of 50 Hz and a peak-to-valley displacement of 15 cm. (The “peak-to- valley” distance is the vertical distance from the farthest positive position to the farthest negative position.) (a) Write the function that describes this wave traveling in the positive x direction. (b) Determine the power being supplied to the string. (c) What is the energy contained in each cycle of the wave?
(a) Sure, I can give you the function that describes this wave traveling in the positive x direction. Let's assume the wave equation is given by y(x, t) = A sin(kx - ωt), where A represents the amplitude, k is the wave number, and ω is the angular frequency.
Given that the segment of the string contains four complete wavelengths, the wavelength (λ) can be calculated as λ = 6 m / 4 = 1.5 m. The wave number (k) can be found by dividing 2π by the wavelength, so k = 2π / λ.
The frequency (f) is given as 50 Hz, which corresponds to the angular frequency (ω) using the relation ω = 2πf.
Now, let's plug in the values we have. A peak-to-valley displacement of 15 cm is equivalent to 0.15 m, so A = 0.15 m. We can also use the given frequency of 50 Hz to calculate ω = 2π(50 Hz) = 100π rad/s.
Using the values of A, k, and ω, the function describing the wave traveling in the positive x direction is y(x, t) = 0.15 sin((2π / 1.5) x - (100π) t).
(b) To determine the power being supplied to the string, we need to know the wave speed (v) of the string. Unfortunately, the wave speed is not provided in the given information. Without that information, it is not possible to calculate the power accurately.
(c) Similarly, to determine the energy contained in each cycle of the wave, we would need the wave speed. Without that information, it is not possible to calculate the energy accurately.
(a) To write the function that describes this wave traveling in the positive x direction, we can use the equation:
y(x, t) = A * sin(kx - ωt + φ)
where:
- A is the amplitude of the wave (peak-to-valley displacement).
- k is the wave number, which relates to the wavelength, λ, and is calculated as k = 2π / λ.
- x is the position along the string.
- ω is the angular frequency, which is calculated as ω = 2πf, where f is the frequency of the wave.
- t is the time.
- φ is the phase constant.
In this case, the peak-to-valley displacement is 15 cm, which is equivalent to 0.15 m. The frequency of the wave is 50 Hz, so the angular frequency ω can be calculated as ω = 2π * 50 = 100π rad/s.
Given that a 6-m segment of the string contains four complete wavelengths, we can calculate the wavelength λ as 6 m / 4 = 1.5 m. Using the wave number equation, k = 2π / λ, we find k = 2π / 1.5 = 4π/3 rad/m.
Therefore, the function that describes this wave traveling in the positive x direction is:
y(x, t) = (0.15) * sin((4π/3)x - (100π)t + φ)
(b) To determine the power being supplied to the string, we need to calculate the energy being transferred per unit time. Power is defined as the rate at which energy is transferred.
The energy of a vibrating string is given by the equation:
E = (1/2)mv^2
where:
- E is the energy.
- m is the mass of the string segment.
- v is the velocity of the string segment.
In this case, the mass of the 6-m string segment is given as 180 g, which is equivalent to 0.18 kg.
The velocity of the string segment can be calculated using the formula:
v = ωA
where:
- v is the velocity.
- ω is the angular frequency.
- A is the amplitude (peak-to-valley displacement).
Substituting the given values, we find:
v = (100π) * (0.15) = 15π m/s
Now, we can calculate the energy E as:
E = (1/2) * (0.18) * (15π)^2
This gives us the energy contained in the 6-m segment of the string.
To find the power, we need to divide the energy by the time it takes for one cycle of the wave. Since the frequency f = 50 Hz, one cycle takes 1/50 s.
So the power being supplied to the string is:
Power = E / (1/50) = 50E
(c) To find the energy contained in each cycle of the wave, we can simply use the energy value we calculated for the 6-m segment of the string.
Therefore, the energy contained in each cycle of the wave is equal to the energy obtained in part (b) divided by the number of cycles in the 6-m segment of the string. As there are four complete wavelengths in the 6-m segment, the number of cycles is 4.
Energy per cycle = Energy / Number of cycles = E / 4