INT One end of a 75-cm-long, 2.5 g guitar string is attached to a spring. The other end is pulled, which stretches the spring. The guitar string's second harmonic occurs at 550 Hz when the spring has been stretched by 5.0 cm. What is the value of the spring constant?

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Accepted Answer

Hey, everyone. So we have a wire of length, 0.8 m and a mass of 10 g that's connected at one end to a spring and at the other end to a clam require one end. You have a plant and the other end and you have a spring. The wire is stretched horizontally to elongate the spring by six centimeters. So the wire is going to stretch the spring. Bye six centimeters point that as 0.6 m. When it's strum, the wire vibrates in its fundamental mode at a frequency of 120 Hertz. So frequency is 120 Hertz and then fundamental mode tells us that the N equals one. We're asked to calculate the spring constant and assume that the spring is at rest. So our multiple choice answers here are a 4.6 times 10 to the two Newton per meter. B 1.9 times 10 to the third newtons per meter. C 6.2 times 10 to the third Newton per meter or D 7.7 times 10 to the third newtons per meter. Ok. So the first thing we can do here is recall Newton's second law, some of the forces, the extraction is equal to mass times acceleration. We know that the spring is at rest, there is no acceleration. So this term goes to zero. And when we look at the forces acting at this point where the spring and the wire are connected, we have tension in the wire and then our spring force in the spring. And so those two forces are going to equal each other. So tension is equal to our spring force. And so we can recall that our spring force is given by K multiplied by delta X where K is our spring constant. And so that's the value that we're ultimately solving for in this crop and tension in the wire or the string is given by new times velocity square. So four the, for the wire, it's essentially um a wire connected at both ends. So we know that velocity is equal to lambda multiplied by the frequency and lambda in turn is equal to two L because we are in our fundamental mode. And then this new term is our linear density. And so by definition, that is simply mass divided by like and so we can pull all of these equations together for tension. And so we get tension equals some new will have mass divided by length multiplied by velocity squared where velocity is two multiply by L multiply by F. And so all of that work, so we can solve that out. And we get four and L squared, F squared divided by L. So one of the owls cancel and we have a 10 of or um L F square. And so now when we plug that back into our Newton Second Law equation, we have four M L F squared equals K delta X. So we will divide both sides by delta X to isolate our spring constant. OK. And so that will be four multiplied by M. So our mass is given to us in the problem as 10 kg, sorry, it's 10 g. So I'm gonna rewrite that as 100.1 kg to keep us in standard units, the length of the wire was given as 0.8 m. And then our frequency is 120 Hertz and that is squared, all of that is divided by delta X which we have said is 0.6 m in standard units. And so plug that into your calculator and we get K is equal to 7.7 times 10 to the third Newton per meter. And so that is the final answer to this problem. And we look at our multiple choice answers and it aligns with answer choice D. So that's all we have for this one. We'll see you in the next video.

INT One end of a 75-cm-long, 2.5 g guitar string is attached to a spring. The other end is pulled, which stretches the spring. The guitar string's second harmonic occurs at 550 Hz when the spring has been stretched by 5.0 cm. What is the value of the spring constant?

Verified Solution

Key Concepts

Harmonics and Frequency

Spring Constant (k)

Tension in a String