Problem 71 - University Physics with Modern Physics

Question

A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of the rock is 3200 kg/m\(^3\). The mass of the wire is small enough that its effect on the tension in the wire can be ignored. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 42.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 28.0 Hz. What is the density of the liquid?

Step-by-Step Solution

Answer

The density of the liquid is approximately 1600 kg/m³.

Step 1: Understand the Problem

We need to find the density of a liquid in which a rock is submerged. The rock affects the frequency of standing waves on a wire when in air versus submerged. The known parameters are the weight and density of the rock, and the fundamental frequencies for waves on the wire in two scenarios - rock in air and submerged.

Step 2: Relate Frequency to Tension

The frequency of standing waves on the wire depends on the tension and the linear mass density of the wire. We can use the formula for the fundamental frequency: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear mass density of the wire.

Step 3: Express Tension in Air

When the rock is in air, the tension in the wire is equal to the weight of the rock. Thus, \( T_{\text{air}} = 164.0 \, \text{N} \). The fundamental frequency is given as 42.0 Hz.

Step 4: Calculate Linear Density in Air

Using the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), rearrange to find \( \mu \) in air: \( \mu = \frac{T}{(2Lf)^2} \). Substitute \( T_{\text{air}} = 164.0 \), \( L = 3.00 \, \text{m} \), and \( f = 42.0 \, \text{Hz} \). This gives \( \mu_{\text{air}} = \frac{164}{(2 \times 3.00 \times 42.0)^2} \).

Step 5: Express Tension in Liquid

When the rock is submerged, the tension in the wire is reduced due to the buoyant force. Let \( T_{\text{liquid}} = T_{\text{air}} - F_{\text{buoyant}} \).

Step 6: Determine Buoyant Force

The buoyant force is the weight of the liquid displaced by the rock, given by \( F_{\text{buoyant}} = \rho_{\text{liquid}} V g \). The volume of the rock \( V = \frac{164.0}{9.81 \times 3200} \). Use this expression in the formula for the buoyant force.

Step 7: Calculate Linear Density in Liquid

Using the new frequency of 28.0 Hz, calculate \( \mu \) while the rock is submerged: \( \mu_{\text{liquid}} = \frac{T_{\text{liquid}}}{(2Lf_{\text{liquid}})^2} \). Substitute values to find \( T_{\text{liquid}} \) in terms of unknown \( \rho_{\text{liquid}} \) and solve for \( \mu_{\text{liquid}} \).

Step 8: Solve for Density of the Liquid

Equate \( \mu_{\text{air}} = \mu_{\text{liquid}} \) since the wire's properties are unchanged. These will yield a relationship in terms of \( \rho_{\text{liquid}} \). Solve for \( \rho_{\text{liquid}} \) using: \[ 164 - \rho_{\text{liquid}} \cdot \frac{164}{9.81 \times 3200} \cdot 9.81 = \left(\frac{42}{28}\right)^2 \times 164 \], then solve for \( \rho_{\text{liquid}} \).

Key Concepts

Buoyant Force

Buoyant force is a significant concept that impacts the behavior of objects submerged in fluids. When an object is immersed in a fluid, such as water or another liquid, it experiences a force that counteracts gravity known as buoyant force. This force arises because of the pressure difference that forms between the top and bottom surfaces of the object.
The buoyant force can be calculated using Archimedes' principle, which states that the force is equal to the weight of the fluid displaced by the object. Mathematically, this is represented as:

• \( F_{\text{buoyant}} = \rho_{\text{liquid}} \times V \times g \)

where \( \rho_{\text{liquid}} \) is the density of the liquid, \( V \) is the volume of the object, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
This force influences how much of an object is below the surface when it floats and also changes the tension in materials attached to the object, such as wires in our exercise.
Understanding buoyant force is crucial for determining changes in tension when an object, like a rock, is submerged in a liquid.

Fundamental Frequency

The fundamental frequency of a system is notably observed in systems that can vibrate and create standing waves, such as a wire with an attached mass. It is the lowest frequency at which a system naturally oscillates, and it forms the basis for higher frequencies or harmonics.
In the context of our exercise, the fundamental frequency of transverse waves on a wire changes when a rock at the end of the wire is submerged in a liquid. This is because the tension in the wire, which influences frequency, is affected by the presence of buoyant force.
Mathematically, the fundamental frequency for transverse waves on a wire can be calculated using:

• \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)

where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear mass density of the wire. Changes in tension, caused by external forces like buoyancy, lead to different calculated fundamental frequencies.

Transverse Waves

Transverse waves are a type of wave where particle displacement is perpendicular to the direction of wave propagation. Picture gently shaking a string to create ripples—this is a simple illustration of transverse waves.
In our problem, transverse waves relate to the vibrations of a wire when a weight is attached and suspended, creating conditions for standing waves. The frequency of these waves depends on factors such as tension in the wire, its mass density, and its length.
Unlike longitudinal waves, where particles move parallel to wave travel (e.g., sound waves in air), transverse waves allow us to visually represent the concept as wave peaks and troughs form along the medium.
It's important to note that, in solid medium applications like a wire, frequency adjustments can occur due to changes in conditions—like submerged objects affecting tension and, subsequently, wave speed.

Weight and Density

Weight and density are key physical properties influencing the behavior of objects in a fluid. Weight is the force acting on an object due to gravity and can be calculated as:

• \( W = m \times g \)

where \( m \) is mass and \( g \) is gravitational acceleration.
Density, on the other hand, is the measure of mass per unit volume, given by:

• \( \rho = \frac{m}{V} \)

In the original exercise, understanding the density of both the rock and the surrounding liquid is crucial. The density of the rock indicates how much fluid it displaces when submerged, which in turn allows us to calculate the buoyant force.
These properties are interconnected, as the rock's weight supports calculations related to tension in the wire both in air and submerged scenarios. By grappling with these concepts, students can solve for unknowns like the density of a liquid based on changes in related physical phenomena.