Problem 53 - University Physics with Modern Physics

Question

Three pieces of string, each of length \(L,\) are joined together end to end, to make a combined string of length 3\(L .\) The first piece of string has mass per unit length \(\mu_{1},\) the second piecehas mass per unit length \(\mu_{2}=4 \mu_{1},\) and the third piece has mass per unit length \(\mu_{3}=\mu_{1} / 4 .(a)\) If the combined string is under tension \(F,\) how much time does it take a transverse wave to travel the entire length 3\(L ?\) Give your answer in terms of \(L, F,\) and \(\mu_{1}\) . (b) Does your answer to part (a) depend on the order in which the three pieces are joined together? Explain.

Step-by-Step Solution

Answer

The total time is \(3.5L\sqrt{\frac{\mu_1}{F}}\), independent of the order of segments.

Step 1: Calculate the Wave Speed for Each Segment

The speed of a wave on a string is given by the formula \( v = \sqrt{\frac{F}{\mu}} \). For each segment, calculate the wave speed:- For the first piece, \( v_1 = \sqrt{\frac{F}{\mu_1}} \).- For the second piece, \( v_2 = \sqrt{\frac{F}{4\mu_1}} \).- For the third piece, \( v_3 = \sqrt{\frac{F}{\frac{\mu_1}{4}}} = \sqrt{4\frac{F}{\mu_1}} = 2\sqrt{\frac{F}{\mu_1}} \).

Step 2: Calculate the Travel Time for Each Segment

The time \( t \) it takes for a wave to travel along a segment of string is \( t = \frac{L}{v} \). Compute the travel time for each piece:- For the first piece, \( t_1 = \frac{L}{\sqrt{\frac{F}{\mu_1}}} = L\sqrt{\frac{\mu_1}{F}} \).- For the second piece, \( t_2 = \frac{L}{\sqrt{\frac{F}{4\mu_1}}} = 2L\sqrt{\frac{\mu_1}{F}} \).- For the third piece, \( t_3 = \frac{L}{2\sqrt{\frac{F}{\mu_1}}} = \frac{L}{2}\sqrt{\frac{\mu_1}{F}} \).

Step 3: Calculate Total Time for the Entire String

Sum the travel times for all three pieces to find the total travel time:\[ t_{\text{total}} = t_1 + t_2 + t_3 = L\sqrt{\frac{\mu_1}{F}} + 2L\sqrt{\frac{\mu_1}{F}} + \frac{L}{2}\sqrt{\frac{\mu_1}{F}} = 3.5L\sqrt{\frac{\mu_1}{F}} \].

Step 4: Evaluate the Dependence on the Order of String Segments

The total time is the sum of individual travel times, which depend only on the individual characteristics of each section, not the order. Thus, the time taken for the wave to travel along the entire length does not depend on the order of the string segments.

Key Concepts

Wave Equation

The wave equation is a fundamental aspect in physics that helps us understand waves travelling in different mediums. Specifically, when analyzing waves on a string, the wave equation connects several critical factors. It tells us about how the tension in a string and its mass per unit length influence the speed at which a wave travels.
The wave speed, denoted as \( v \), is calculated using the formula:

• \( v = \sqrt{\frac{F}{\mu}} \)

where \( F \) is the tension in the string, and \( \mu \) is the mass per unit length. This relationship shows that wave speed increases with greater tension and decreases with higher mass per unit length.
Understanding this principle is crucial when solving problems involving wave dynamics on a string, as it underscores the interplay between physical properties and the behavior of waves.

Tension in Physics

Tension is a force applied along a string or rope, helping to transmit waves. It's essential in wave behavior because it directly affects the speed of the wave's travel. In simple terms, the higher the tension in a string, the faster a wave can travel along it. This is because increased tension reduces the amount of sag or slack, providing a more direct route for the energy to travel.
In our problem, tension \( F \) plays a key role in determining the speed of the wave for each segment of string. Regardless of the mass changes across the string segments, the tension remains constant. Thus, it is a critical factor in calculating wave speed using the wave equation formula mentioned above. Without adequate tension, waves would not effectively propagate along the string.

Mass per Unit Length

Mass per unit length, represented as \( \mu \), is another vital concept in understanding waves on a string. It describes the distribution of mass along the string and significantly impacts the wave speed.

• Lower mass per unit length results in faster wave speeds, as there is less resistance to wave motion.
• Higher mass per unit length means slower wave speeds, as the waves must work against more inertia.

In the exercise, different segments of the string have varying mass per unit lengths—\( \mu_1, 4\mu_1, \frac{\mu_1}{4} \). This variation affects how quickly a wave can travel through each segment. For instance, in the lighter segment (mass per unit length of \( \frac{\mu_1}{4} \)), the wave travels fastest because it has the least mass to move. Conversely, in the heaviest segment (mass per unit length of \( 4\mu_1 \)), the wave travels slowest due to increased inertia.
Thus, mass per unit length is a crucial factor that, along with tension, determines the wave's speed across a string.