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(1) The A string on a violin has a fundamental frequency of 440 Hz. The length of the vibrating portion is $32 \mathrm{cm},$ and it has a mass of 0.35 $\mathrm{g}$ . Under what tension must the string be placed?

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$87 \mathrm{N}$

Physics 101 Mechanics

Chapter 16

Sound

Periodic Motion

Mechanical Waves

Sound and Hearing

University of Michigan - Ann Arbor

Simon Fraser University

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.

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A violin string 30.0 $\mat…

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burned 40 hurts. The length of the vibrating string portion is 32 centimeters and it has a mass of 0.35 grams. Under what tension must the stream be placed? Okay, so we need to figure out the tension. So I wrote down were given that the frequency of the fundamental notice 440 hurts. The length is 32 verdict. Two meters. There's 100 centimeters in one meter. So that 0.32 meters and the masses 35 grams which I convert two kilograms to keep everything in s I u nits. So there are 1000 grams and kilograms. So 0.35 grams 0.35 times 10 to the minus three kilograms. All right, so let's use our equation for this fundamental frequency that's equal to the velocity divided by two times the length of the strength. Okay, but the velocity is equal to the square root of the force from the tension which will call f C t divided by Nass, divided by a length. Okay, so we can plug that in for V and we're gonna have frequency is equal to one over to L times the square root force of attention divided by the mass, divided by the length. Right, That mass There. Okay, so now we know everything in the equation except for the force of the tension, which is what we're trying to find so we can rearrange his equation to solve for the force of the tension. So F gets multiplied by two out. So we're gonna end up with force attention. On one side, the frequency gets multiplied by two times the length, So that gets moved over frequency, then two going okay. Now we need to square everything. Okay? I mean, to square this and then we need to multiply by the length over the mess now. Okay. So if we, uh, if we plug all of that into our equation here, take here. Yeah, that's right. Okay, so this is divided by, so we'll just kind of leave it like this so that this is divided by mass over length. Still right. Okay. So we need to multiply my length. I'm sorry. Multiplied by mass and divide by length, so multiplied by mass divide by weight. Yeah. There we go. Okay. So this will be gone, and you'll have. This is equal to F square times four times L times. Okay, well, we have all those values above there. You plug all of those in and you'll find this is equal to 87 in the units on this. Our new things. What is your solution?

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