When mass M is tied to the bottom end of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 200 Hz. Adding an additional 1.0 kg to the hanging mass increases the second-harmonic frequency to 245 Hz. What is M?

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So this problem is dealing with standing waves in a string that is fixed on both ends. So let's see what it's asking us consider an in extensible and massless string of length L suspended from the top of a roof. A block of mass B is firmly attached to the string's lower end. When the string is wiggled. A third harmonic standing wave of frequency 250 Hertz is set up on the string. If an additional mass of 1.5 kg is added to the lower end, the third harmonic frequency is shifted to 285 Hertz, calculate the block's mass. Our multiple choice answers here are a three kg B five kg, C eight kg or D 11 kg. OK. So the first thing I'm going to do here is draw a free body diagram. So we have our string, we have a block hanging down from our string. This is our initial block of mass feed. So initially the forces acting on this block, our tension in the positive right direction and weight in the negative right direction. And so when we go to do our, um when we can recall Newton's second law F equals mass multiplied by acceleration. We can sum the forces in the Y direction. So we will have tension minus weight and then there is no acceleration. Uh The block is just hanging there. So that is zero. So our tension is equal to our weight. And then in turn, we can recall that weight is equal to mass multiplied by gravity. And so this mass here is what we are in the end trying to solve for the problem is talking about frequency and the third harmonic frequency. And so when we have tension and frequency, there are two equations that we can recall that kind of tie this together. So frequency of the string is equal to the, the velocity divided by lamb of the wavelength. And in turn velocity is equal to the square root of tension divided by mu. They also tell us that we are in the third harmonic frequency. So we can recall for the third harmonic or lambda is equal to our wavelength is equal to two L divided by three. And so we have two different scenarios here. When we're looking at the frequency, you have our initial um scenario where we just have the mass of the block and a frequency of 250 Hertz. And so we'll call that scenario one. And then our second scenario will be after we add that additional weight. So our initial frequency is equal to, we have BS substitute in the uh square root of T divided by mu multiplied by Lambda. I'm sorry, divided by Linda. So that would be multiplied by three divided by two. And we have already established that tension is equal to mass multiplied by gravity. So we can rewrite this as three divided by two L multiplied by you swear rude. A mass of the block multiplied by gravity divided by mute. When we square both sides, they are left with nine divided by four L squared multiplied by the mass of the block multiplied by gravity all divided by you. So this new term is interesting. It's the uh linear density of the string. And so in both of our scenarios, that term is going to be the same linear density of the string doesn't change. And so we can rearrange this equation to solve for you. And then that will help us when we are looking at our second scenario. So let's be U equals nine divided by four L squared multiplied by M B multiplied by G all divided by our initial read when they swear. OK. So let's just, that's just gonna hang out there for a minute. And now we're gonna look at our second scenario. This is after we add the additional 1.5 kg. So we have our final three received is again, going to be equal to three divided by two L multiplied by the square root. Now, instead of just mass of the block, it's mass of the block plus 1.5 kg. All of that multiplied by gravity divided by mu. And so when we go to square both sides of this equation, you are left with nine divided by or else squared multiplied by M B plus 1.5 kilograms multiplied by gravity divided by mute our new term we solved or in from the first equation we are going to substitute in here. So this looks like nine divided by four elsewhere multiplied by M B plus 1.5 kg. That quantity multiplied by G and then multiplied by the inverse of mu. So that will be four L squared I sweared divided by nine M B. So right off the bat, we can simplify the nines cancel, the force, cancel our lengths, cancel. And so we simplify this. Now it looks like F F our final frequency sweared is equal to our mass plus 1. kg times gravity times our initial frequency squared divided by mass. Sorry. It said the mass of the blocks multiplied by a graph. If we multiply it through the gravity terms and divide both sides by our initial frequency, we are left with F F squared divided by F I squared is equal to mass multiplied by gravity plus 1.5 multiplied by gravity all divided by mass multiplied by. And that simplifies to just one bus, 1.5 divided by nuts. And so now we are getting very close to this math term. We're almost there. That is the end goal here. Um So I appreciate you all staying sticking with me so far, we just have a few more steps. So we have our final frequency squared divided by our initial frequency squared minus one is equal to 1.5 divided by mass. And so our last step is multiplying both sides by mass and then dividing both sides like this equation. And so that leaves us with mass is equal to 1.5, multiplied by one, divided by R final frequency squared, divided by our initial frequency squared minus one. So it's a little, a little messy. But now we have our mass isolated on the left side of the equation and we have all of our known values. So sorry. So at some point I I dropped our units on the 1. that's still kilograms. So let's put this back in there. And so our mass is equal to 1.5 kg times one divided by 285 Hertz squared, divided by 250 Hertz squared minus one. And we plug that into our calculator and we get 5.0. And so that is the answer to this problem. We go back up to multiple choice answers and that aligns with answer choice B, this is a tricky one. Thanks for sticking with me through it. We'll see you in the next video.

When mass M is tied to the bottom end of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 200 Hz. Adding an additional 1.0 kg to the hanging mass increases the second-harmonic frequency to 245 Hz. What is M?

Verified Solution

Key Concepts

Second-Harmonic Frequency

Tension in a Wire

Mass and Frequency Relationship