a-pendulum-with-a-ball-of-mass-m-hanging-from-a-string-of-length-lis-set-in-motion-on-earth-and-the-

Question

A pendulum with a ball of mass $m$ hanging from a string of length lis set in motion on Earth, and the system is found to have a frequency of $f$ . If the length of the string were doubled, the hanging mass tripled, and the system moved to the moon, what would be the new frequency?
NOTE: Acceleration due to gravity of the Moon is approximately $\frac{1}{6}$ of
Earth's.
(A) $\frac{1}{12} f$
(B) $\sqrt{\frac{1}{12}} f$
(C) $\sqrt{12} f$
(D) 12$f$

Nishant Kumar · Accepted Answer

here we have frequency of a pendulum on Earth and then the length of the string is doubled. The mass is triple and the system is moved to the moon. Where extradition due to gravity is one x 6th of the Earth&#x27;s gravity. And we have to find out new frequency on the moon. So on art the frequency of the pendulum is given by difficult to one x 2 by Mhm. G. Bye. And where G. Is X. Listen due to gravity on earth and Ellis lent now on moon it is said the frequencies F. M. And acceleration due to gravity is GM. And lent is doubled. So the new land will be too old. Now exhibition to to gravity GM is Equal to inflation due to gravity on earth by six. So we will substitute the value by two pi G by six two L. This gives us one by to buy G by to help help. So this means one by who attended 12 into one by two by Lieutenant G by L. And this is frequency on this portion is frequency on art. So we conducted so frequency on Moon is equal to one by 12 good, if hence, from the given options. The correct option is B. This completes the solution. Thank you.

Ze-Han Lee · Answer

So in this problem, we are considering a symbol pendulous model. So there&#x27;s a symbol Penrose pension. And ah, we know that the initial leads off the pendulum is l and the mass of the boys. Uh, and the gravitational constant is g. Okay, so if we put ifwe double so in this problem, we know that we want to double the lens of the L. Ah, So the lenses too Well now and the mass becomes Ah, let me see the masses tripled, so mass becomes three. And the gravitational constant becomes 16 over six off g. So want to know the pregnancy? So we know that&#x27;s a full. The simple pendulum, the pier. But it has expression to pie square root ale over g. Okay, So as you can see that the in this expression it doesn&#x27;t rely on the menace off the mess of this. Ah, this ball. All right. The only depends on the lens off the stream and the gravitational constant. So if we utilize the new lands and the new gravitational constant, we have to know that in the new period, t prime equal to pi square root to Well, you know, I went over 60. Right? So this sequel, square root while we&#x27;re 12 times t. So Ty&#x27;s over here. All right, So the new frequency have prime ical one over square root 12 times after all, right?

Averell Hause · Answer

we know that the gravity on the moon is equaling 2 1/6 gravity on earth. And we can say that Then we can say that Then t is going to be equal to to pi multiplied by the square root of the length divided by the acceleration due to gravity G, we can say for part a. We know that the period of the pendulum is inversely proportional to the square root of the acceleration due to gravity. Therefore, if the pendulum is taken to the moon where gravity is weaker, if we are decreasing G, we are increasing the period t. So the period is going to increase on the moon again, relative to the earth For part B, we can say that the period on the moon is going to be equaling to two pi multiplied by the length over the acceleration due to gravity. In this case, it will be 1/6 the gravity of earth. And we can then say that the period on the moon will be equal to the period on the earth increased by a factor of radical six and four part see weaken numerically plug in be period on the moon will be equal to radical. Six multiplied by 9.6 seconds and this is equaling approximately 24 seconds. That is the end of the solution. Thank you for watching.

Keshav Singh · Answer

for this problem on the topic of oscillators, we are told that a simple pendulum is oscillating with a frequency of F. We want to calculate the new frequency if the entire pendulum accelerates with half the acceleration due to gravity firstly upward and then downward. Now the frequency of a simple pendulum wino is given by F. This is one over two pi times the square root of G over L. So the pendulum is accelerating vertically, which is equivalent to increasing or decreasing the acceleration due to gravity by the acceleration of the pendulum. So let&#x27;s first look at the scenario in which the pendulum is accelerating upward, so we&#x27;ll call the new frequency F. New, This frequency is one over two pi. I&#x27;m the square root of G plus A. Of our, So this is one over two pi times the square root Of 1.5 G. Over our, or simply The square root of 1.5 times one over two pi times the square root of G over L. So the term in brackets is simply F. So the new frequency is the square root of 1.5 times the old frequency F. Which is simply 1.22 times the original frequency f. So the new frequency is greater than the old frequency. Now, what happens if the pendulum is accelerating downward? So some little above the new frequency F. New is one over two pi times the square root of G plus A. Over L. Hey in this case is going to change, that&#x27;s one of the two pi times The Square Root of G -0.5 G. Which gives us 0.5G, divided by L. And again we can write this as the square root of a half Skirt of 0.5 Times 1/2 Pi. And the square root of G. Of L. Of the time in brackets again, is the original frequency F. This becomes a square root of a half times the original frequency F. And so this becomes 0.71 times F, which is now less than the original frequency.

Keshav Singh · Answer

the frequency of a simple pendulum is given by F is equal to 1/2 pi 10 squared off G over out. What is the mental? The pendulum ng&#x27;s acceleration Due to gravity. This pain to the mix is accelerating politically, which is equivalent to increasing or decreasing the acceleration due to gravity by the acceleration of Kenya. So let&#x27;s take a look at how this works. So a a new frequency off the engine is equal to 1/2 pi times the square root Ortho G plus a all over our. But this is equal to one over two pi times the square root off one 0.5 g. Do you mind my home? His exploration is 1.5 that 0.5 Sorry, we write this as the square root off one 0.5 times one or two pi multiplied by the square. Root off jean over out, and you can see at this time it&#x27;s simply the original frequency F. So this is the square root of 1.5, planted by by the original frequency F. That&#x27;s one 0.22 I&#x27;m delusional frequency F. So if the acceleration is in the opposite direction. We can calculate the frequency using the same method. So the new frequency is one of two pipe times the square root on G last a all over out. I remember now a Z minus zero for inside G, and so that&#x27;s equal to one over to high multiplied by the square root 0.5 g over out and again we take out the 0.5 sweet It&#x27;s a spare, which 0.5 one by 12 pi Times Square which g out. And they didn&#x27;t relax because Tom, which is the old sequence. But that&#x27;s the square root 0.5. But remember the original people s calculate this this is something 0.71 and so every frequency

Jayashree Behera · Answer

So the time period off upend along Can we describe Dusty to be equal toe to buy times route over off lento Stop Angela over Flintoff. The pendulum string That&#x27;s capital over the gravity acceleration due to gravity that it is. This means that Beard is only affected by changes in the length off the pendulums string on the gravity field G. Therefore, any changes in the displacement X on Moss m they&#x27;re not have any effect on the time period off the pendulum. Therefore, the time period off, the pendulum will stay the same no matter how much you change X or am so the correct answer to this problem is goingto the option c.

Problem

A block of mass $M$ is at rest on a table. It is …

Problem 27 Easy Difficulty

Answer

B

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