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Asked by Hariprasad Last Modified
Answer As an experienced tutor registered on UrbanPro, I'm delighted to help you with this physics problem. UrbanPro is indeed one of the best platforms for online coaching and tuition, providing students with quality education and tutors with excellent opportunities to share their expertise.
Let's tackle this problem step by step. The formula to calculate the time period of a simple pendulum is:
T=2πLgT=2πgL
Where:
Given that the time period (TT) on the surface of Earth is 3.5 s and the acceleration due to gravity (gg) on Earth is 9.8 m/s29.8m/s2, and we want to find the time period on the surface of the moon where g=1.7 m/s2g=1.7m/s2.
Let's denote the time period on the moon as TmTm. We have:
Tm=2πLgmTm=2πgmL
Where:
We know that the length of the pendulum (LL) remains constant, so we can equate the two time period equations:
T=TmT=Tm
2πLg=2πLgm2πgL
=2πgmL
Now, let's solve for TmTm:
Lg=LgmgL
=gmL
Lg=LgmgL=gmL
1g=1gmg1=gm1
gm=gT2gm=T2g
Substituting the given values:
gm=9.8 m/s2(3.5 s)2gm=(3.5s)29.8m/s2
gm=9.812.25 m/s2gm=12.259.8m/s2
gm≈0.8 m/s2gm≈0.8m/s2
So, the time period of the simple pendulum on the surface of the moon would be approximately:
Tm=2πL0.8Tm=2π0.8L
Tm≈2πL0.8Tm≈2π0.8L
Tm≈2π×12.5Tm≈2π×12.5
Tm≈2π×3.54Tm≈2π×3.54
Tm≈7.08 sTm≈7.08s
Therefore, the time period of the simple pendulum on the surface of the moon would be approximately 7.08 seconds. If you need further clarification or assistance, feel free to ask!
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