Problem 50 - University Physics with Modern Physics

Question

A 1750-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (\(A\) and \(B\)), each 1.25 m long and weighing 0.290 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? Which pulse arrives first? (Ignore the effect of the weight of the wires on the tension in the wires.)

Step-by-Step Solution

Answer

The pulse through wire B arrives 0.00234 s before the pulse through wire A.

Step 1: Identify the problem

We need to find the time delay between pulses transmitted to the ceiling through the two wires, A and B, when plucked at the same time. We know the mass of the beam and the length and weight of wires.

Step 2: Determine the position of the center of gravity

The center of gravity is one-third the way from one end (wire A) along the beam, which implies that two-thirds of the beam's weight contributes to wire B and one-third to wire A.

Step 3: Calculate tension in each wire

Using the beam's total weight (1750 N), the tension in wire A \[ T_A = rac{1}{3} imes 1750 ext{ N} = 583.33 ext{ N} \]And the tension in wire B \[ T_B = rac{2}{3} imes 1750 ext{ N} = 1166.67 ext{ N} \]

Step 4: Determine wave speed in each wire

The speed of wave (\( v \)) in a wire is determined by\[ v = \sqrt{\frac{T}{\mu}} \]where\( \mu = \frac{m}{L} \)is the mass per unit length. Using the weight of the wire (0.290 N) to find its mass,\( m = 0.290 ext{ N} / 9.8 ext{ m/s}^2 = 0.0296 ext{ kg} \).Mass per unit length\( \mu = 0.0296 ext{ kg} / 1.25 ext{ m} = 0.02368 ext{ kg/m} \).Then calculate:\[ v_A = \sqrt{\frac{583.33}{0.02368}} \approx 156.7 ext{ m/s} \]\[ v_B = \sqrt{\frac{1166.67}{0.02368}} \approx 221.5 ext{ m/s} \]

Step 5: Determine arrival time of waves

Using the speeds, the time taken for the wave to travel each wire is\[ t_A = \frac{1.25 ext{ m}}{156.7 ext{ m/s}} \approx 0.00798 ext{ s} \]\[ t_B = \frac{1.25 ext{ m}}{221.5 ext{ m/s}} \approx 0.00564 ext{ s} \]

Step 6: Calculate the time delay and determine which pulse arrives first

The time delay is given by the difference in arrival times,\[ \Delta t = t_A - t_B \approx 0.00798 ext{ s} - 0.00564 ext{ s} = 0.00234 ext{ s} \].The pulse through wire B arrives first.

Key Concepts

Tension in Wires

Tension in wires plays a crucial role in wave propagation. When a beam is suspended by wires, the tension is determined by the weight distributed on each wire. In our example, the beam weighs 1750 N with the center of gravity located one-third from wire A.
This means two-thirds of the weight is supported by wire B, resulting in varying tensions:

• For wire A, the tension is calculated as one-third of the beam's weight: \( T_A = \frac{1}{3} \times 1750 \text{ N} = 583.33 \text{ N} \).
• For wire B, the tension is two-thirds of the beam's weight: \( T_B = \frac{2}{3} \times 1750 \text{ N} = 1166.67 \text{ N} \).

Higher tension in a wire allows faster wave propagation. This principle is essential in engineering and physics applications where tension significantly affects the behavior of structures and transmitted waves.

Center of Gravity

The center of gravity refers to the point at which an object's mass is evenly distributed. In the context of a suspended beam, it affects how weight is shared between supporting wires.
In our scenario, the center of gravity is one-third of the distance from the end where wire A is attached. This placement indicates:

• One-third of the beam's weight is supported by wire A.
• Two-thirds of the beam's weight is supported by wire B.

Understanding the center of gravity is vital in determining the force distribution across a structure, influencing engineering designs for stability and balance. It is crucial for ensuring structures like bridges and buildings can withstand applied forces without tipping.

Mass per Unit Length

Mass per unit length ( \( \mu \) ) represents the mass distributed over a given length of material. It is critical in calculating wave speed in wires. To find \( \mu \) , use the formula:\[ \mu = \frac{m}{L} \] where \( m \) is the mass and \( L \) the length of the wire.In our example, the wire weighs 0.290 N, which gives it a mass of:

• \( m = \frac{0.290 \text{ N}}{9.8 \text{ m/s}^2} = 0.0296 \text{ kg} \).
• With a length of 1.25 m, the mass per unit length is: \( \mu = \frac{0.0296 \text{ kg}}{1.25 \text{ m}} = 0.02368 \text{ kg/m} \).

The smaller the mass per unit length, the higher the wave speed, illustrating why lighter, longer wires are effective in faster wave transmission. This concept is widely applied in crafting musical instruments, enhancing designs for better acoustics.