π times square root of its original size L 1 end g and also then it's gonna have actually some new period T 2 i beg your pardon is 2π times square root of its brand-new length together 2 over g and we are told the L 2 is 2 time L 1. Therefore we'll substitute in 2L 1 in ar of l 2 here and this works out come 2π time square source L 1 end g times square root of 2. Now all of this is T 1 and also so we can replace this through T 1 here and I guess we should write the in red to display that it's a replacement and also we see that T 2 climate is square root 2 times T 1. So once doubling the length, the period increases through a variable of square root 2 i m sorry is about 1.4. In component (b) we are supposed to mean that size two is 5 percent much less than length one so a reduction of 5 percent method multiply l 1 through a aspect of 0.9500. So we plug in 0.9500L 1 in ar of together 2 and this works out come 0.9500 times T 1 and also if you take it the square root of 0.950, it's 0.975 for this reason the period decreases by 2.5 percent due to the fact that this is 2.5 much less than 100. So 2.5 percent as soon as decreasing the size by 5 percent.">

You are watching: What is the effect on the period of a pendulum if you double its length? This is university Physics Answers through Shaun Dychko. A pendulum has actually an initial duration of 2π time square root of its original size L 1 end g and then it"s gonna have some new period T 2 i m sorry is 2π times square root of its brand-new length together 2 end g and we room told the L 2 is 2 times L 1. Therefore we"ll substitute in 2L 1 in place of l 2 here and also this works out come 2π times square source L 1 over g time square source of 2. Now every one of this is T 1 and so we deserve to replace this with T 1 here and also I assumption: v we must write that in red to show that it"s a replacement and also we check out that T 2 then is square root 2 times T 1. So as soon as doubling the length, the duration increases through a aspect of square root 2 which is about 1.4. In part (b) we are claimed to suppose that length two is 5 percent much less than length one therefore a reduction of 5 percent means multiply l 1 through a factor of 0.9500. For this reason we plugin 0.9500L 1 in place of together 2 and also this works out to 0.9500 time T 1 and also if you take the square root of 0.950, it"s 0.975 for this reason the period decreases by 2.5 percent since this is 2.5 less than 100. So 2.5 percent when decreasing the size by 5 percent.
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