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\(\mathrm{CP}\) A \(1750-\mathrm{N}\) irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires \((A\) and \(B),\) each 1.25 \(\mathrm{m}\) long and weighing 0.360 \(\mathrm{N}\) . 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? (Neglect the effect of the weight of the wires on the tension in the wires.)

Short Answer

Expert verified
Wire A's pulse arrives first with a delay of 0.0089 s.

Step by step solution

01

Understand the System

The system consists of a 1750 N beam suspended by two vertical wires, A and B, each 1.25 m long and weighing 0.360 N. The beam's center of gravity is one-third of the way along the beam from wire A.
02

Find Tension in Wires

Since the beam is in equilibrium, the sum of forces and torques must be zero. The center of gravity is one-third of the way from wire A, meaning wire A carries more of the weight. Using torque equilibrium, calculate the tension in wire A as \(T_A = \frac{2}{3} \times 1750 \text{ N} = 1166.67 \text{ N}\) and tension in wire B as \(T_B = \frac{1}{3} \times 1750 \text{ N} = 583.33 \text{ N}\).
03

Determine Wave Speeds

The wave speed in each wire depends on the tension and the linear mass density \(\mu\). The linear mass density is found using the weight and length: \(\mu = \frac{0.360 \text{ N}}{1.25 \text{ m}} = 0.288 \text{ kg/m}\). The wave speed is given by \(v = \sqrt{\frac{T}{\mu}}\).
04

Calculate Wave Speed for Each Wire

For wire A: \(v_A = \sqrt{\frac{1166.67}{0.288}} \approx 63.75 \text{ m/s}\). For wire B: \(v_B = \sqrt{\frac{583.33}{0.288}} \approx 43.87 \text{ m/s}\).
05

Calculate Time of Pulse Travel

To find the time taken for a pulse to travel the length of each wire, use \( t = \frac{L}{v} \).For wire A: \(t_A = \frac{1.25}{63.75} \approx 0.0196 \text{ seconds}\).For wire B: \(t_B = \frac{1.25}{43.87} \approx 0.0285 \text{ seconds}\).
06

Determine Time Delay and Which Pulse Arrives First

The time delay between the pulses is \( t_B - t_A = 0.0285 - 0.0196 = 0.0089 \text{ seconds}\). The pulse in wire A arrives first, as it has a higher wave speed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension in Wires
Tension in wires refers to the force that stretches the wire when a load is applied. In this problem, the beam is supported by two wires, so knowing the tension helps in understanding how the load is distributed between them. Tension is important because it affects the wire's behavior, just like stretching a rubber band changes its feel.
When determining the tension in a system like this, equilibrium plays a crucial role. Since the beam is stationary, the forces and torques (rotational forces) must balance each other out—like a seesaw that doesn't tip. The center of gravity of the beam is closer to wire A, indicating wire A holds more weight.
To calculate the tension, consider the distribution of the beam's weight. Wire A carries about two-thirds of the beam's weight, calculated as \( T_A = \frac{2}{3} \times 1750 \text{ N} \approx 1166.67 \text{ N} \). Conversely, wire B supports the remaining one-third, calculated as \( T_B = \frac{1}{3} \times 1750 \text{ N} \approx 583.33 \text{ N} \). This balance helps keep the beam steady and horizontally aligned.
Wave Speed Calculation
Wave speed in a wire can be determined by the tension and the wire's linear mass density (mass per unit length). The formula used is \( v = \sqrt{\frac{T}{\mu}} \) where \( T \) is the tension and \( \mu \) is the linear mass density. This relationship shows that wave speed is higher when tension increases or the wire is lighter.
In this exercise, each wire's linear mass density is derived from its weight and length: \( \mu = \frac{0.360 \text{ N}}{1.25 \text{ m}} = 0.288 \text{ kg/m} \). This indicates how heavy the wire is compared to its length, much like how some strings are thicker than others.
For wave speed in wire A, with higher tension, the calculation is \( v_A = \sqrt{\frac{1166.67}{0.288}} \approx 63.75 \text{ m/s} \). For wire B, with less tension, it is \( v_B = \sqrt{\frac{583.33}{0.288}} \approx 43.87 \text{ m/s} \). This demonstrates that wire A, being under more tension, can transmit vibrations faster.
Equilibrium of Forces
Equilibrium of forces is a fundamental principle in physics where all acting forces balance each other, resulting in a system that does not change - it remains still or moves consistently. In the case provided, the beam is not accelerating or rotating, indicating equilibrium. The net force and the net torque both equal zero.
Forces in equilibrium mean that any additional or reduced force could disturb the balance, much like slight pressure on a balanced scale tips it. Here, the force exerted by the beam's weight is counteracted by the tensions in the wires. Torque equilibrium ensures the beam does not rotate. Wire A, being closer to the center of gravity, experiences more tension, maintaining stability.
Such equilibrium principles are widely applicable, from constructing bridges to using a lever. Balancing forces ensures that objects stay stable and function correctly, avoiding-collapse or unwanted motion.

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Most popular questions from this chapter

A horizontal wire is stretched with a tension of 94.0 \(\mathrm{N}\) and the speed of transverse waves for the wire is 492 \(\mathrm{m} / \mathrm{s}\) . What must the amplitude of a traveling wave of frequency 69.0 \(\mathrm{Hz}\) be in order for the average power carried by the wave to be 0.365 \(\mathrm{w}\) ?

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