Author Topic: GMA Chapter 4A- direct variation (Read 1962 times) Tweet Share

dot223 · « **on:** December 30, 2013, 04:16:43 pm »

Question 8- the time taken for one complete oscillation of a pendulum is called it's period. The period (T) of a pendulum varies directly with the square root of the length (L) of the pendulum. A pendulum of length 60cm has a period of 1.55 seconds. Find the period of a pendulum that's is one and a half times as long.
I got approx 2.45 seconds though the answer is 1.898
Which is given when converting cm into m. Why does that have to be done? How come you can't get the answer by leaving the units as they are?

lzxnl · « **Reply #1 on:** December 30, 2013, 04:22:28 pm »

Quote from: dodic123 on December 30, 2013, 04:16:43 pm

Question 8- the time taken for one complete oscillation of a pendulum is called it's period. The period (T) of a pendulum varies directly with the square root of the length (L) of the pendulum. A pendulum of length 60cm has a period of 1.55 seconds. Find the period of a pendulum that's is one and a half times as long.
I got approx 2.45 seconds though the answer is 1.898
Which is given when converting cm into m. Why does that have to be done? How come you can't get the answer by leaving the units as they are?

So... $T_1^2=kL_1$
Scale length by factor 1.5 => $T_2^2=kL_2$
Divide the two equations
$\frac{T_2^2}{T_1^2} = \frac{L_2}{L_1} = 1.5$
$\frac{T_2}{T_1} = \sqrt{1.5}$
$T_2 = \sqrt{1.5} T_1 = 1.898 s$

dot223 · « **Reply #2 on:** January 02, 2014, 09:56:33 am »

thanks!!

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Author Topic: GMA Chapter 4A- direct variation (Read 1962 times) Tweet Share

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GMA Chapter 4A- direct variation

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