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Tuning a cello. A cellist tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 $\mathrm{m}$ long and has a mass of 14.4 $\mathrm{g} .$(a) With what tension must she stretch that portion of the string? (b) What percentage increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from $\mathrm{C}$ to $\mathrm{D} ?$

a) 147.8 $\mathrm{N}$b) 26$\%$

Physics 101 Mechanics

Chapter 12

Mechanical Waves and Sound

Periodic Motion

Mechanical Waves

Sound and Hearing

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

08:15

In physics, sound is a vibration that typically propagates as an audible wave of pressure, through a transmission medium such as a gas, liquid or solid. In human physiology and psychology, sound is the reception of such waves and their perception by the brain. Humans can only hear sound waves as distinct pitches when the frequency lies between about 20 Hz and 20 kHz. Sound above 20 kHz is known as ultrasound and has different physical properties from sound below 20 kHz. Sound waves below 20 Hz are called infrasound. Different species have different hearing ranges. In terms of frequency, the range of ultrasound, infrasound and other upper limits is called the ultrasound.

04:49

In physics, a traveling wave is a wave that propogates without a constant shape, but rather one that changes shape as it moves. In other words, its shape changes as a function of time.

06:40

Tuning an Instrument. A mu…

03:37

A cellist tunes the C-stri…

01:21

05:51

A musician tunes the C-str…

03:39

While attempting to tune t…

08:45

04:04

01:31

$\bullet$ Tuning a violin.…

So we have a fundamental frequency of 65.4 hertz. We have the length of the string of 0.0 of 0.8 00 meters. And so for party, we know that the mass of the string equals 14.4 grams or weaken. Say 0.144 kilograms. So we can say that the fundamental frequency equals thie. Ah, velocity divided by two times the length. And we know that the velocity equals the square root of the force. Tension divided by Mew Mew Here is the linear density. So essentially, um, Mu equals the mass divided by the length so we can solve for forced tea and say forced tension equals Aah V squared em over l and we can solve so rather refit me should solve for V and say that this is going to be two times the length of the fundamental frequency squared times m divided by Al. Now, this is simply going to be equal to four times the fundamental frequency squared times the length times the mass. And now we can solve, so forced tension will be equal to four times 65 0.4 squared times 0.800 times 0.144 and this is equally 148 units. So this will be our final answer for part a. The attention in the string so that she can hit the fundamental frequency of 65.4 Hertz. Now for Part B, we have a new fundamental frequency. So the F one of new is going to be equal to 73.4 hertz, and we need to find the new force tension. So the new force tension simply going to be equal? Teo, 73.4, hurts, um, divided by 65.4 hertz squared times the original tension of 148 Newtons and this is going to equal 186 Newton's. This is on ly this. We can only do this because the force tension is directly proportional to the frequency squared. So essentially you can make a relationship and find the new force tension. We're hitting a frequency that is higher than the original frequency. So the fundamental frequency actually increases in this case, which means that the tension must also increase in this case and they're asking for the percentage increase as well. So I have 186 Newton's minus 148 noons divided by 148 times, 100% and this is giving us 25.7% increase. Intention. So this will be one answer, and this will be the increase that that is the end of the solution. Thank you for watching.

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