A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m

A large rock that weighs 164.0 N is suspended from the...

A large rock that weighs $164.0 \mathrm{~N}$ is suspended from the lower end of a thin wire that is $3.00 \mathrm{~m}$ long. The density of the rock is $3200 \mathrm{~kg} / \mathrm{m}^{3} .$ 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 $42.0 \mathrm{~Hz}$. 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 $28.0 \mathrm{~Hz}$. What is the density of the liquid?

Key Concepts

Tension in a Suspended System

In a system where an object is suspended by a wire, the tension in the wire is generally equal to the net force acting on the object due to gravity and any other vertical forces. When additional forces such as a buoyant force are present, the effective tension in the wire decreases by the magnitude of this upward force, altering the mechanical behavior of the system.

Buoyant Force and Archimedes’ Principle

When an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the displaced fluid. This principle, known as Archimedes’ principle, is critical for determining how immersion modifies forces in a system. It directly influences the effective tension in the wire by reducing the net downward force acting on the attached object.

Standing Wave Frequency on a String

The fundamental frequency of standing waves on a stretched wire or string is determined by the length of the string, the tension in the string, and the linear mass density of the string. This frequency is described by the relation f = (1/(2L)) * sqrt(T/?), which links the physical parameters of the string to its oscillatory behavior.

Relationship Between Tension and Wave Speed

The speed of transverse waves traveling along a string is derived from the square root of the ratio of tension to the linear mass density (v = sqrt(T/?)). Changes in tension thus directly affect the speed at which waves travel along the string and, in turn, the frequencies of standing waves formed on the string.

Mass, Density, and Volume Relationships

The concepts of mass, density, and volume are interconnected by the relation density = mass/volume. In problems involving buoyancy, determining the volume of the object is essential because it is used to calculate the buoyant force. Knowing the object's density and mass allows for the computation of its volume, which then facilitates analysis of how much fluid is displaced when the object is submerged.

Transcript

00:01 We have a rock from a long, basically, string, which is three meters long, and the rock weighs 164 newtons.

00:10 And when it's hanging here, the string has a frequency of 42 hertz.

00:17 And that's the frequency for transverse standing waves.

00:21 And then we then submerge the rock in liquid.

00:26 I drew it here as water, but we don't know what the liquid is.

00:29 And then when you do that, the frequency changes to 28 hertz.

00:34 And if you recall, the frequency is going to change because the velocity is going to change here.

00:40 So recall one of our fundamental equations, let's set up the problem is v is square root of f over a mu, just to give us an idea why the frequency changes.

00:51 And the goal here is to find the density of the liquid.

00:56 So i also recall that v is going to be equal to lay, lambda f, v is going to be lambda f.

01:11 And the buoyant force, which we'll call b, is the row of the liquid times the volume of the object displacing it.

01:22 So call it vr for volume of rock times g.

01:28 And then also recall that the sum of all the forces in equilibrium have to cancel out to be zero.

01:39 So we can calculate.

01:40 The wavelength here is going to be equal to 2l we're in the fundamental frequency so that's going to be six meters in air the velocity is going to be equal to lambda f which is 42 hertz times six centimeters which is two fifty two meters per second so the mass per unit length the string is going to be f over v squared which is one sixty four newtons over two 252 meters per second squared, which is 0 .002583 kilograms per meter.

02:31 So in the liquid, so this is for air, in the liquid, we're going to have a different velocity.

02:38 Got a conserve space.

02:41 Liquid.

02:42 Our velocity is going to be equal to lambda f, which is 28 hertz times 6 meters...

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