The pendulum in a grandfather clock is made of brass and keeps...

Question

The pendulum in a grandfather clock is made of brass and keeps perfect time at 17°C. How much time is gained or lost in a year if the clock is kept at 26°C? (Assume the frequency dependence on length for a simple pendulum applies; see Chapter 14.)

Accepted Answer

Welcome back. Everyone in this problem. The pendulum in a desk clock is made of aluminum and synchronized to give perfect time at 15 °C. If the temperature changes, the clock loses its synchronization as its frequency depends on its length. Just like a simple pendulum, determine the time gained or lost in a month if the clock is kept at 25 °C. A hint is that the coefficient of linear expansion for aluminum alpha is approximately 23 multiplied by 10 to the negative sixth per degree Celsius. A says five minutes are gained. B five minutes are lost. C 3.8 minutes are gained and D says 3.8 minutes are lost. Now, let's first make note of the information we have here. OK. So far, we know that at a perfect time, we have our temperature of 15 °C and we want to know the time gained or lost if the clock is kept at 25 °C. So that means our change in temperature is the difference between 25 and 15 °C, which is 10 °C. We also know that our coefficient of linear expansion for aluminum alpha is 23 multiplied by 10 to the negative sixth per degree Celsius. Now, how is this going to help us figure out the time that's gained our last in a month? What do we know? Well, there are two relationships that are really important here. First, let's talk a little bit about the relationship between temperature and length. In our problem statement, it tells us as the temperature changes. Ok, the clock loses its synchronization as its frequency depends on its length. So the temperature is proportional to the length. As a matter of fact, we can recall, OK, that the length for this new temperature is going to be equal to the initial length L knot, the, the initial length L knot, this is the length at 15 °C multiplied by one plus the coefficient of linear expansion multiplied by the change in temperature. OK. So from this relationship, we can tell that our length for our pendulum is directly proportional to the change in temperature as the temperature increases, the length will increase and as it decreases. So will the length decrease? Now, what other relationship is important? Well, where it tells us that the frequency depends on the length. OK. Now there's a relationship between the period and the length of the pendulum. OK. Based on the formula that tells us that the period T is equal to two pi multiplied by the square root of the length of the pendulum divided by G the acceleration due to gravity. So what this tells us, OK. For example is that for our initial temperature T equals two pi multiplied by the square of LK not divided by G. OK. Well, for our new, for our new temperature, we'll have a new period that's going to be equal to two pi multiplied by the square root of our new length L divided by G. So here we can use this information to help us to find the time gained that are lost because the clock will lose time since the period has increased because the temperature has increased. So at 15 °C, it would show T seconds. But at 25 °C, it would take T prime seconds to complete the cycle, but it would only show two seconds. So that means some time is being lost. Now to figure out what time that is being lost, we'll need to find the change in our period. OK. So what do we know? Well, if we try to find the change in our period, the fractional change, that's a change in our period compared to the original period, it's going to be equal to T prime minus T divided by T. Now we have expressions for each of these. So if we go ahead and substitute, that's going to be two pi multiplied by the square root of L divided by G minus two pi multiplied by the square of L not divided by G or original length divided by two pi root L not divided by G. Now, if we look at these, look at our equation here. OK. We notice that we can cancel two pi from all the terms. And we can also cancel G when we do that, we're left with root L minus root L knot divided by root L knot. But this is good because remember we have an expression for L in terms of L knot. So if we rewrite L in terms of LNO, really, then it's going to be the square root of LNO multiplied by one plus alpha multiplied by the change in temperature minus root on. OK. Divided by root or not. No, again, we're in a similar situation, we can cancel ill, not from all of our terms. And no, when we go ahead and simplify, we get our fractional change to be root one plus alpha multiplied by the change in temperature minus one. Let's substitute our values. So now that's going to be the skirt of one plus alpha 23 multiplied by 10 to the negative sixth per degree Celsius multiplied by a change in temperature 10 °C. And all of that is going to subtract one. Now, when we do that, we get our fractional change to be 1.1499 multiplied by 10 to the negative fourth. So this is going to be the fractional change in the period. So now that means over a month. Ok, over a month and we can take a month to be 30 days. Then the time that's lost is going to be equal to a fractional change. 1.1499 multiplied by 10 to the negative fourth, multiplied by the number of seconds in a day. 86,400 ok, seconds per day multiplied by the number of days in a month. So that's 30 days per month. And now when we work the so, ok, we do get the time lost to be approximately equal to 298.05 seconds. If we convert that to minutes, it's gonna be approximately 4.968 minutes. And if we write that to two significant figures, it's approximately five minutes. So increasing our temperature to 25 °C makes our clock lose approximately five minutes in a month. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

The pendulum in a grandfather clock is made of brass and keeps perfect time at 17°C. How much time is gained or lost in a year if the clock is kept at 26°C? (Assume the frequency dependence on length for a simple pendulum applies; see Chapter 14.)

Key Concepts

Thermal Expansion

Pendulum Frequency and Length Relationship

Time Period of a Pendulum

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