Temperature and the period of a pendulum For oscillations of s...

Question

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation:

T = 2π√(L/g),

where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,

dL/du = kL.

Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT/2.

Accepted Answer

Welcome back, everyone. A spherical balloon's volume changes with temperature at a rate directly proportional to its volume. The volume B of the balloon can be expressed as V equals 4/3 by our cube, where R is the radius of the balloon and R is a function of temperature T. The rate of change of the radius with respect to temperature is given by DR divided by D T equals KR. Where k is a proportionality constant, determine the rate in terms of V at which the volume changes with respect to the temperature. Now, we're given the volume function which is 4/3 by our cubed. What we know is that volume. is a function of radius, and radius is a function of temperature. So what we can tell is that volume is a composite function V of R of T. So when we went to identify The rate of change of volume, which is DV divided by DT in in terms of temperature, we need to apply the chain rule. First of all, we have to differentiate the volume function. With respect to radius. And then we are going to multiply. By the derivative. Of DR divided by DT. The rate of change of radius with respect to temperature. So now let's identify each derivative. Let's begin with the DB divided by DR. That's the derivative of 4/3 pi multiplied by our cubed. We can apply the power rule, which gives us 4/3 by multiplied by 3 are squared, right? So that's 4 pi are squared. From here, let's identify DR divided by DT which is already given to us, right? We know that DR divided by DT is equal to K multiplied by R. Combining those in our chain rule, we can show that the rate of change of volume. With respect to temperature is equal to 4 pi R2. Multiplied by KR. This gives us 4 pi KRQ. But we want to express it in terms of. Volume and not radius. We know that volume equals 4/3 by R cubed, so we can just solve for R cubed, multiplying both sides by 3/4, so it gives us. R cubed equals 3/4 volume. Divided by pi. And essentially what we can do is write volume in the numerator, so we get 3 B. And 5 in the denominator, so we get 3 B divided by 4 pi. And then we can substitute this into our expression DV divided by DT. Which gives us 45k. Multiplied by 3 B divided by 4 pi. Let's notice that we can cancel out 4 pie. And we have our final expression, which is 3 kV. Divided by 1, or basically 3 kV. Well then, we have expressed the rate of change of volume with respect to change in temperature in terms of volume, that's 3 KV. Thank you for watching.

Key Concepts

Differentiation

Chain Rule

Proportionality and Linear Differential Equations

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