154,140 views
39 votes
39 votes
A pendulum on a grandfather clock

is supposed to oscillate once every
2.00 s, but actually oscillates once
every 1.99 s. How much must you
increase its length to correct its
period to 2.00 s?
(Unit = m)

User Profet
by
2.8k points

2 Answers

15 votes
15 votes

Answer:.0099

Explanation:accelus

User Arthur Kazemi
by
2.8k points
21 votes
21 votes

you have to increase 0.01 meters the length

Step-by-step explanation

The period of a pendulum is given by:


T=2\text{ }\pi\sqrt[]{(L)/(g)}

where T is the period, L is the length , a g is the the acceleration of the gravity (9.8 m per square second)

s

Step 1

find the length, for period= 2 s

Let


T_1=2\text{ seconds}

with this value, we can find the length

replace, and isolate L


\begin{gathered} T=2\text{ }\pi\sqrt[]{(L)/(g)} \\ 2=2\text{ }\pi\sqrt[]{(L)/(g)} \\ \text{divide both sides by 2}\pi \\ (2)/(2\pi)=\frac{2\text{ }\pi}{2\pi}\sqrt[]{(L)/(g)} \\ (1)/(\pi)=\sqrt[]{(L)/(g)} \\ ((1)/(\pi))^2=(\sqrt[]{(L)/(g)})^2 \\ (1)/(\pi^2)=(L)/(g) \\ \text{Multiply both sides by g} \\ (1)/(\pi^2)\cdot g\cdot g=(L)/(g) \\ (g)/(\pi^2)=L \end{gathered}

so, when the period is 2.00 s the length of the pendulum is


\begin{gathered} L_1=(g)/(\pi^2) \\ L_1=(9.8)/(\pi^2) \\ L_1=0.9929475997\text{ m} \end{gathered}

Step 2

now, for Period = 1.99 s

Let


T=1.99\text{ s}

replace and solve for L


\begin{gathered} T=2\text{ }\pi\sqrt[]{(L)/(g)} \\ 1.99=2\text{ }\pi\sqrt[]{(L)/(g)} \\ \text{divide both sides by 2}\pi \\ (1.99)/(2\pi)=\frac{2\text{ }\pi}{2\pi}\sqrt[]{(L)/(g)} \\ (1.99)/(2\pi)=\sqrt[]{(L)/(g)} \\ ((1.99)/(2\pi))^2=(\sqrt[]{(L)/(g)})^2 \\ \\ 0.1003105048=(L)/(g) \\ \text{Multiply both sides by g} \\ 0.1003105048\cdot g=(L)/(g)\cdot g \\ 0.1003105048\cdot g=L \\ L=0.1003105048\cdot9.8 \\ L_2=0.983049474\text{ m} \end{gathered}

so, when the period is 1.99 s , the length is 0.9830429474 m

Step 3

finally, to know how much you must increase the length, subtract L2 from L1

so


\begin{gathered} L_1-L_2=0.9929475997\text{ m-}0.9830429474m \\ L_1-L_2\approx0.99-0.98 \\ L_1-L_2\approx0.01 \end{gathered}

therefore, the make the pendulum osscilates with period of 2 seconds, you have to increase 0.01 meters the length

I hope this helps you

User JorgeeFG
by
2.9k points