1. A fisherman's scale stretches 3.6 cm when a 2.5-kg fish hangs from it.
What will be the frequency of oscillation if the fish is pulled down 2.7 cm more and released so that it oscillates up and down?
2. Two trains emit 518 Hz whistles. One train is stationary. The conductor on the stationary train hears a 3.1 Hz beat frequency when the other train approaches.
What is the speed of the moving train?
3.When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged.
If the unfingered length of the string is l= 65 cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 2^{1/12
I know that K= 680 N/m
1. Yes, k = Mg/X = 681 N/m
The frquency of oscillation does not depend upon how much farther the scale is pulled down. It is just
f = [1/(2 pi)]*sqrt(k/m)
2. The frequency of the whistle from the oncoming train must be 521.1 Hz
I don't have time to do your third question
We would prefer you post one question at a time. You will probably get answers quicker that way. It is also impportant to show your own work
The doppler shift ratio is
3.1/f = V/(340 m/s)
so V = 340*3.1/518 = 2.0 m/s
There is a more exact doppler shift formula for moving sources, but that is accurate enough.
Ty. I will keep that in mind next time.
If anyone could help with the last one, I would greatly appreciate it.
1. To find the frequency of oscillation, we can use the formula:
f = 1 / T
Where f is the frequency and T is the period. The period is the time it takes for one complete oscillation.
First, let's find the spring constant of the fisherman's scale. We can use Hooke's Law:
F = k * Δx
Where F is the force, k is the spring constant, and Δx is the displacement. In this case, the force is the weight of the fish, which is given as 2.5 kg (converted to newtons by multiplying by 9.8 m/s^2). The displacement is 3.6 cm (converted to meters by dividing by 100).
Using Hooke's Law, we can rearrange the formula to solve for k:
k = F / Δx
Next, we want to find the new displacement when the fish is pulled down 2.7 cm more. The total displacement will be 3.6 cm + 2.7 cm = 6.3 cm (converted to meters by dividing by 100).
Now, we can find the new spring force using the same formula:
F = k * Δx
Where F is the new force, k is the spring constant we calculated earlier, and Δx is the new displacement.
Finally, we can find the period of oscillation using the formula:
T = 2π * √(m / F)
Where T is the period, m is the mass, and F is the force.
Plug in the values we have:
m = 2.5 kg
F = new force calculated earlier
The period T is the time for one complete oscillation, so the frequency f is 1 / T.
2. To find the speed of the moving train, we can use the formula for the Doppler effect:
f' = f * (v + v₀) / (v - v₀)
Where f' is the observed frequency, f is the emitted frequency, v is the speed of sound, v₀ is the speed of the stationary train, and v is the speed of the moving train.
We are given that the observed beat frequency is 3.1 Hz and the emitted frequency is 518 Hz.
Since the conductor on the stationary train hears the beat frequency, the observed frequency (f') is the sum of the emitted frequency and the beat frequency:
f' = f + Δf = 518 Hz + 3.1 Hz
We know the speed of sound is approximately 343 m/s.
Rearranging the Doppler effect equation, we can solve for the speed of the moving train:
v = (v₀ * (f' - f)) / (f' + f)
Plug in the values we have:
v₀ = speed of the stationary train
f' = observed frequency
f = emitted frequency
3. To determine the positions x of the first six frets on the guitar string, we need to know the corresponding frequencies of the notes produced by each fret and use the relationship between frequency and length of the vibrating portion of the string.
The formula that relates frequency (f) and length (l) of a vibrating string is:
f = v / (2 * l)
Where f is the frequency, l is the length, and v is the wave speed.
In this case, the length of the unfingered string is given as l = 65 cm (converted to meters by dividing by 100).
The wave speed v is determined by the tension and mass per unit length of the string. Since these remain unchanged as indicated in the question, we can consider v as a constant for our calculations.
To determine the positions x of the first six frets, we can use the ratio of frequencies of neighboring notes from the equally tempered chromatic scale, which is given as 2^(1/12).
Starting with the unfingered length, we can calculate the frequency for each fret using the formula:
fret frequency = unfingered frequency * (2^(1/12))^fret_number
Where fret frequency is the frequency of the note produced by the specific fret, unfingered frequency is the frequency of the unfingered open string, fret_number is the fret number (1 for the first fret, 2 for the second fret, and so on).
We can calculate the frequencies for the first six frets using this formula. Then, to find the position x for each fret, we can rearrange the formula for frequency:
l = v / (2 * f)
and solve for length:
l = v / (2 * (fret frequency))
Substituting the known values for length and frequency, we can solve for the position x.
Repeat this process for each of the six frets to determine their respective positions.