A cord of length 1.5 m is fixed at both ends. Its mass per unit length is 1.2 g / m and the tens

A cord of length 1.5 m is fixed at both ends. Its mass per...

A cord of length $1.5 \mathrm{m}$ is fixed at both ends. Its mass per unit length is $1.2 \mathrm{g} / \mathrm{m}$ and the tension is $12 \mathrm{N} .$ (a) What is the frequency of the fundamental oscillation? (b) What tension is required if the $n=3$ mode has a frequency of $0.50 \mathrm{kHz} ?$

Key Concepts

Tension and Frequency Relationship

The tension in the string plays a crucial role in determining the frequency of vibration. Increasing the tension increases the wave speed, which in turn increases the natural frequencies of oscillation. This relationship is pivotal when adjusting or controlling the pitch of a string in musical instruments or when calibrating devices that rely on string vibrations.

Harmonics and Modes

When a string fixed at both ends vibrates, it does so in discrete modes or harmonics. The nth harmonic has a frequency that is an integer multiple of the fundamental frequency, given by f? = n f? or equivalently f? = (n/(2L))?(T/?). This concept is important for understanding the harmonic series and how higher frequency modes are produced by the same string under different conditions or configurations.

Fundamental Frequency

The frequency of the fundamental oscillation, also known as the first harmonic, is the lowest frequency at which a string fixed at both ends vibrates. It is given by the formula f? = (1/(2L))?(T/?), where L is the length of the string. This concept highlights the direct relationship between physical parameters such as tension, mass per unit length, and string length with the vibrating frequency.

Wave Speed on a String

The wave speed on a stretched string is determined by the tension and the mass per unit length of the wire or cord. This relationship is expressed by the formula v = ?(T/?), where T is the tension in newtons and ? is the linear mass density in kilograms per meter. This concept is fundamental in understanding how the properties of a string affect the propagation of waves along it.

Transcript

00:01 So this question gives us the length of a string of 1 .5 meters, the linear mass density of 1 .2 grams per meter, and the tension in the string of 12 neutrons.

00:11 The first part of the question asks us to work out the fundamental frequency, which we're going to call f1.

00:20 And we know from the textbook equation that that is equal to 1 over 2l, where l is our length, multiplied by the square root of t over mu where t is our tension mu's our linear mass density so putting in some values that's equal to one over making sure everything is in s iunits two and then our l 1 .5 meters then multiplied by the square root of our tension which is 12 newtons over our linear mass density which we need to convert to kilograms per meter so that's just simply 1 .2 times 10 to the minus 3 kilograms per meter remembering the neutins on the top of our fraction as well and just making a bit more room here once we plug those numbers into the calculator that comes out as 33 hertz the two two significant figures which is the answer quoted in the textbook.

01:35 So the second part of the question asks us to work out, well it gives us an n value of three and it gives us a frequency at the third highest resonance of 0 .5 kilohertz.

01:59 And we are asked to work out the tension in the string given this information.

02:04 So that equation that we used at the start for the fundamental frequency, we can sort of use an analog of that to describe what's going on with this third resonant frequency here.

02:20 So we can say that f3, instead of f1, is equal to the only thing that changes is we have three times by our fundamental frequency...

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