Let's tackle your questions one at a time, starting with the tuning forks. The statement "The 1st fork is an octave of the last" is crucial for understanding the relationship between the frequencies of the forks. In music, an octave means that one frequency is double the frequency of another. So, if we denote the frequency of the last fork as \( f \), the frequency of the first fork would be \( 2f \). Now, let's break down the problem step by step.
Understanding the Tuning Forks
We have a set of 25 tuning forks arranged in decreasing frequency. The first fork has a frequency that is double that of the last fork. This means:
- Let the frequency of the last fork be \( f \).
- Then, the frequency of the first fork is \( 2f \).
Next, we know that each fork produces 3 beats with the succeeding one. The beat frequency is the absolute difference between the frequencies of two tuning forks. Therefore, if the frequency of the first fork is \( 2f \) and the second fork is \( f_2 \), we can express the beat frequency as:
Beat frequency: \( |2f - f_2| = 3 \)
Finding the Frequencies
Since the forks are arranged in decreasing order, we can express the frequency of each fork in terms of the last fork's frequency. The frequency of the \( n \)-th fork can be represented as:
Frequency of the n-th fork: \( f_n = 2f - (n-1)d \)
Here, \( d \) is the common difference in frequency between successive forks. We know that:
- For the first fork (n=1): \( f_1 = 2f \)
- For the second fork (n=2): \( f_2 = 2f - d \)
- For the third fork (n=3): \( f_3 = 2f - 2d \)
- ... and so on until the last fork (n=25): \( f_{25} = 2f - 24d = f \)
From the last fork's frequency, we can set up the equation:
Equation: \( 2f - 24d = f \)
Rearranging gives us:
Solving for d: \( f = 24d \) or \( d = \frac{f}{24} \)
Now, substituting \( d \) back into the beat frequency equation for the first and second forks:
We have:
Beat frequency: \( |2f - (2f - d)| = 3 \)
This simplifies to:
Equation: \( d = 3 \)
Now substituting \( d \) back into \( d = \frac{f}{24} \):
Final Calculation: \( 3 = \frac{f}{24} \) leads to \( f = 72 \) Hz.
Thus, the frequencies are:
- First fork: \( 2f = 144 \) Hz
- Last fork: \( f = 72 \) Hz
Analyzing the Rope and Wave Propagation
Now, let's move on to the second question regarding the rope. When a uniform rope of length \( L \) and mass \( m \) is whirled in a horizontal circle, we can ignore the effects of gravity. The key here is to find the time required for a transverse wave to travel from one end of the rope to the other.
Wave Speed in the Rope
The speed of a transverse wave on a rope is determined by the tension in the rope and its mass per unit length. The formula for wave speed \( v \) is given by:
Wave speed: \( v = \sqrt{\frac{T}{\mu}} \)
Where:
- \( T \) is the tension in the rope.
- \( \mu \) is the mass per unit length, calculated as \( \mu = \frac{m}{L} \).
When the rope is whirled, the tension \( T \) can be expressed in terms of the angular velocity \( \omega \) and the radius of the circular path. If we assume the rope is being whirled at one end, the tension can be approximated as:
Tension: \( T = m \cdot \omega^2 \cdot r \), where \( r \) is the effective radius of the circular motion.
Calculating the Time for Wave Travel
Now, substituting \( T \) and \( \mu \) into the wave speed equation:
Wave speed: \( v = \sqrt{\frac{m \cdot \omega^2 \cdot r}{\frac{m}{L}}} = \sqrt{L \cdot \omega^2 \cdot r} \)
To find the time \( t \) it takes for the wave to travel the length \( L \) of the rope, we use the formula:
Time: \( t = \frac{L}{v} \)
Substituting the expression for \( v \) gives us:
Final Time Calculation: \( t = \frac{L}{\sqrt{L \cdot \omega^2 \cdot r}} = \frac{\sqrt{L}}{\omega \sqrt{r}} \)
In summary, the time required for a transverse wave to travel from one end of the rope to the other depends on the length of the rope, the angular velocity, and the effective radius of the circular motion. This relationship highlights the interplay between tension, mass, and wave propagation in a physical medium.
