# a pendulum-driven clock gains 5.00 s/day, what fractional change

in pendulum length must be made for it to keep perfect time?

## T = 2π√(L/g)

or, in terms of L:

L = g(T/2π)²

To cut down on writing, let's define a constant k = g/(2π)². Then the above is:

L = kT²

So a small change in length (dL) is related to a small change in period (dT) like so (via derviative):

dL = 2kTdt

But notice (from previous equation) that kT = L/T. Substitute:

dL = 2(L/T)dt

The fractional change in pendulum length is (by definition) (dL)/L. By the above equation, this is:

(dL)/L = 2(dt)/T

That is, fractional change in pendulum length is twice the fractional change in period. Now, we know the clock is ticking too fast, and we want to slow it down by 5.90s per day. That is accomplished by increasing the period (making it tick slower) by a fraction of (5.90 s per day). That is:

(dt)/T = 5.90s per day = 5.90s / 86400s = 6.83×10^-5

So the fractional change in pendulum length would be:

(dL)/L = 2(dt)/T = 2(6.83×10^-5) = 1.36×10^-4

## Well, if a pendulum-driven clock gains 5.00 s/day, it's safe to say that it has a "time management" problem, don't you think? It's probably thinking, "Why rush when you can just take your sweet time?" Now, to make it keep perfect time, we need to figure out the fractional change in pendulum length required.

Let's see...a pendulum's period (T) is affected by its length (L). The relationship is expressed as T = 2π√(L/g), where g is the acceleration due to gravity. If we want the clock to gain no time, we need to change the period to keep it constant.

Since the period is given by T = 24 hours, we can calculate the original pendulum length (L₀) using T₀ = 2π√(L₀/g), where T₀ is the original period.

Now, the new period (T₁) will be T₀ + (5.00 s/day). So we can write T₁ = 24 hours + (5.00 s/day).

To find the new pendulum length (L₁), we can rearrange T = 2π√(L/g) to get L = (T/2π)²g.

So, L₁ = (T₁/2π)²g, and L₀ = (T₀/2π)²g.

To find the fractional change in pendulum length (ΔL), we can subtract L₀ from L₁ and divide by L₀, like this:

ΔL = (L₁ - L₀)/L₀.

By plugging in the values we calculated, we can find the answer. However, since I can't do math, I suggest using a calculator or consulting an expert to crunch the numbers for the precise value of the fractional change in pendulum length needed. Good luck, and hopefully, the clock will learn to manage its time a bit better!

## To find the fractional change in pendulum length that must be made for the clock to keep perfect time, we can use the equation for the period of a pendulum:

T = 2π√(l/g)

where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity.

Let's assume that the current length of the pendulum is l₁. Since the clock gains 5.00 seconds per day, the actual period (T₂) is greater than the ideal period (T₁) by 5.00 seconds:

T₂ = T₁ + 5.00 seconds

Using the equation for the period, we have:

2π√(l₂/g) = 2π√(l₁/g) + 5.00 seconds

Simplifying, we can cancel out the 2π:

√(l₂/g) = √(l₁/g) + 5.00 seconds

Squaring both sides of the equation, we get:

l₂/g = (l₁/g) + 10(l₁/g) + 25

Multiplying through by g, we have:

l₂ = l₁ + 10l₁ + 25g

Since we want to find the fractional change in length, we can express the change in length as a fraction of the original length:

Δl = l₂ - l₁

Δl/ l₁ = l₁ + 10l₁ + 25g - l₁

Δl/ l₁ = 10l₁ + 25g

Δl/ l₁ = 10(1 + 2.5(g/l₁))

Therefore, the fractional change in pendulum length that must be made for the clock to keep perfect time is 10(1 + 2.5(g/l₁)).

Note that the value of g is approximately 9.81 m/s².

## To determine the fractional change in pendulum length needed for the clock to keep perfect time, we can use the formula for the period of a pendulum. The period (T) of a pendulum is given by:

T = 2π√(L/g)

Where L is the length of the pendulum and g is the acceleration due to gravity. Since we want to find the fractional change in pendulum length, we'll express the period as a function of the length, and then differentiate with respect to length.

So, let's start by rearranging the formula for the period:

T = 2π√(L/g)

(T/2π)² = L/g

L = (gT²)/(4π²)

Next, we differentiate L with respect to T:

dL/dT = (g * 2T)/(4π²)

dL/dT = (gT)/(2π²)

Now, we can substitute the values given in the problem to calculate the fractional change in pendulum length. We know that the clock gains 5.00 s/day, which means the period of the pendulum increases by 5.00 seconds per day. Converting this to seconds per second, we have 5.00 s/day * (1 day/24 hours) * (1 hour/60 minutes) * (1 minute/60 seconds) = 5.78704 x 10^(-5) s/s.

So, the fractional change in pendulum length is:

ΔL/L = (dL/dT) * (ΔT/T)

ΔL/L = (gT)/(2π²) * (5.78704 x 10^(-5) s/s) / T

ΔL/L = (g * 5.78704 x 10^(-5))/(2π²)

Now, we need the value for acceleration due to gravity (g). Assuming it's approximately 9.81 m/s², we can substitute it into the equation:

ΔL/L = (9.81 m/s² * 5.78704 x 10^(-5))/(2π²)

Calculating this expression will give you the fractional change in pendulum length required for the clock to keep perfect time.