A 350 g mass on a 45-cm-long string is released at an angle of 4.5° from vertical. It has a damping constant of 0.010 kg/s. After 25 s, (a) how many oscillations has it completed and (b) what fraction of the initial energy has been lost?

Question

Diagram of a pendulum with a 35 cm string at a 6° angle from vertical.

Accepted Answer

Hey, everyone in this problem, a block having a mass of 500 g is tied to a thread with a length of 35 centimeters. The block is gonna be pulled sidewards such that it makes an angle of six degrees with the vertical. Then it's gonna be released and the block is gonna oscillate with damping. OK? And the damping constant is 0.005 kg per second. Now, we are asked to determine the number of oscillations after 30 seconds and also the fraction of energy lost. So we're showing our diagram here. We have our block tied to our thread that's 35 centimeters long making the six degree angle with the vertical. And we're given four answer choices. OK. Each of them have a different combination of the number of oscillations and the energy lost. And we'll come back to those as we work through this problem. Now, let's start with the first part. OK. The number of oscillations after 30 seconds. So the number of oscillations OK? In 30 seconds is gonna be equal to the number of oscillations per second multiplied by the number of seconds, which is gonna be 30 hm. All right. Now, the number of oscillations per second recall that this is just the frequency. So what we're really doing is we're gonna take the frequency F, we're gonna multiply it by the number of seconds which is 30 seconds to get the number of oscillations in 30 seconds. Now, how can we calculate? F? OK. Will recall that we have the angular frequency omega which is equal to two pi F um which is sorry, which is equal to the square root of K divided by M. And regularly we would leave it right here. But in this case, we have damping. OK. Our oscillations are damping. So we have to consider that. And so instead of just the square root of K divided by M, we have the square root of K divided by M minus B squared, divided by four M squared A to account for that damping. Now, this tells us that our frequency F that we're trying to find is gonna be one divided by two pi multiplied by the square root of K divided by M minus B squared, divided by four M squared. OK. So looking at this, do we have everything we need? Do we know all of these spirits? OK. Well, we know be, we know the mass M but we don't know we are given the value of K. What we do know though is that K divided by M is equal to omega knots squared. OK. That angular frequency if we didn't have damp. So this is gonna be one divided by two pi multiplied by the square root of Omega knott squared minus B squared divided by four M squared. OK. And we're just working with this equation to try to write it in terms of some variables that we have. OK. Well, Omega knot squared, OK. I recall that Omega NOTT squared can also be written as the square root of G divided by L. We have one divided by two pi multiplied by the square root of G divided by L minus B squared divided by four M squared. And now we are getting somewhere, OK? We know G the acceleration due to gravity, we know L the length of that string we were given at B is a damping constant which we were given and we have the mass. So now we have everything we need to calculate this frequency and it's just a matter of substituting in these values. Now, let's write out all of these values convert our into our standard units first. OK. So the acceleration due to gravity 9.8 m per second squared, that one's OK. The length of this string we're told is 35 centimeters. We wanna convert that into our standard unit of meters. We're gonna multiply by 1 m divided by 100 centimeters. OK? We're essentially dividing by 100 we get 0.35 m. What else do we need? We need this damping constant B which we're told is 0.00 5 kg per second. And we need our mask, which we're told is 500 g. Again, converting to our standard unit, we multiply by 1 kg divided by 1000 g because we know there are 1000 g in every kilogram. The unit of gram divides out, we're essentially dividing by 1000 and we get 0.5 kg. So substituting these into our equation for the frequency F one divided by two pi multiplied by the square root of 9.8 m per second squared, divided by 0.35 m minus 0.005 kg per second. All squared divided by four multiplied by 0.5 kg squared. Let's look at our units first, we have meters per second squared divided by meters. It's gonna give us a unit of per second squared in the first term under the square root. The next term we have kilogram per second all squared. So that's kilograms squared per second squared divided by kilograms squared. So again, we're getting per second squared, when we take the square root, we get per second, which is gonna be equivalent to a Hertz. And if we work all of this out on our calculator, we get that the frequency F is equal to 0.84217 Hertz. All right. So we're gonna put a blue box around that So we don't lose track of it. And we needed that frequency to calculate the number of oscillations that occur in 30 seconds. OK. So again, the number of oscillations in 30 seconds is equal to the frequency F multiplied by 30 seconds. This is equal to 0.84217 Hertz multiplied by 30 seconds, which gives us 25.265 oscillations. And that is the first part of this problem. We found about 25 oscillations in 30 seconds. If we go up to our answer choices, we can already eliminate option B because it has 30 oscillations for this portion of the problem. And we can eliminate option C because it has 10 oscillations. OK? Option A and D both have 25 oscillations, which is what we found to be correct. Now, I wanna move to the second part of this problem and the second part is asking for the fraction of energy lost. Now, just like when we calculated the frequency F, we have to be a little bit careful because we have damping here with damping. That means that our energy E is gonna decay over time. So our energy is gonna be a function of time, then it's gonna be equal to eno multiplied by E to the exponent negative B T divided by M. What is eno Well, that's gonna be that initial energy and in this case, it's gonna be gravitational potential energy. Let me go back to our diagram. OK. When that block is pulled to the left, it's still being held. It's not moving, there's no kinetic energy, but we do have some gravitational potential energy because it has some height. Now, we have to figure out what that height is. What we can imagine is that if we were to let this go and the string was lying straight vertically, the entire height would be 35 centimeters. OK. The length of that strength when we raise it up, it's gonna be higher than that point. And so this difference, it's gonna be the height H we're looking for. All right. So that height is gonna be 35 centimeters minus the right hand side of this triangle we've made. And the right hand side of this triangle we've made is gonna be related through the cosine of the angle because it's the adjacent side. So let's get back to our equation here and write out what we've just talked about. So again, we have gravitational potential energy ug recall that this is equal to MG multiplied by H and in this case, H knot, our mass is 0.5 kg acceleration due to gravity 9.8 m per second squared. And this height H is gonna be that total length of the strength, 0.35 m minus that length of the triangle we drew which was 0.35 m multiplied by cosine of six degrees and that is gonna give you that leftover part at the bottom, the amount that we've raised that block by. All right. Now, if we work all of this out, we're gonna have 0.009395 we have kilogram meter per second squared multiplied by meter, which is equivalent to a duel. Ok. So that's eno in order to find the fraction lost, we need to find the energy at 30 seconds as well. So the energy at 30 seconds is gonna be given by the equation we've already written. We have eno 0.009395 Jews multiplied by E to the exponent negative B 0.005 kg per second multiplied by the time 30 seconds divided by the mass 0.5 kg. And this is gonna give us an energy after 30 seconds of 0.00696 jules. Ok? So we have our eot we have our energy E at 30 seconds. We want to find the fraction of energy lost. OK? So the fraction lost. Let me write it on the right hand side here. So we can leave everything else up fraction lost. Well, this is gonna be that initial energy in minus the energy at 30 seconds. OK? That's gonna be the amount lost. We want the fraction lost. So we're gonna divide that by that initial energy email, we have 0.009395 minus 0.00696 all divided by 0.009395. Now, our units are jewels for all of these values. And so that unit is gonna divide out and we're gonna be left with no units. We have 0.25 92 and that is gonna be that fraction lost. All right. So going back up to our answer choices now that we've answered the second part and we can see that these are in percentages instead of decimals. OK? To convert from the decimal, we found to the percent we want to multiply by 100. And if we do that, we can see that option. A is the correct answer. We have 25 oscillations in those 30 seconds and we're gonna lose 25.9% of our energy. Thanks everyone for watching. I hope this video helped see you in the next one.

A 350 g mass on a 45-cm-long string is released at an angle of 4.5° from vertical. It has a damping constant of 0.010 kg/s. After 25 s, (a) how many oscillations has it completed and (b) what fraction of the initial energy has been lost?

Verified Solution

Key Concepts

Simple Harmonic Motion (SHM)

Damping

Energy Conservation in Oscillatory Systems