Verified Solution • 2025–26 Session moderate

Q13.15: NCERT Physics - Part 2 OSCILLATIONS 13.15 Solution: The acceleration due to gravity on the surface of moon is 1.7 m …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often invert the gravity ratio, writing $T_{moon}=T_{earth}\sqrt{g_{moon}/g_{earth}}$, which would make the pendulum swing faster on the Moon. Remember that for a fixed length $l$, $T\propto1/\sqrt{g}$, so the correct scaling is $T_{moon}=T_{earth}\sqrt{g_{earth}/g_{moon}}$.

The acceleration due to gravity on the surface of moon is 1.7 m s__. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s__)

Step-by-Step Verified Solution

**Answer:** Tmoon=8.4T_{moon} = 8.4 s **Step-by-Step Explanation:** On earth: T=2πl/g⇒l=T2g4π2=(3.5)2×9.84π2=3.04T = 2\pi\sqrt{l/g} \Rightarrow l = \frac{T^2 g}{4\pi^2} = \frac{(3.5)^2 \times 9.8}{4\pi^2} = 3.04 m. On moon: T′=2πl/g′=2π3.041.7=8.4T' = 2\pi\sqrt{l/g'} = 2\pi\sqrt{\frac{3.04}{1.7}} = 8.4 s [citation:14].

All Questions in this Chapter

Q13.1 Q13.2 Q13.3 Q13.4 Q13.5 Q13.6 Q13.7 Q13.8 Q13.9 Q13.10 Q13.11 Q13.12 Q13.13 Q13.14 Q13.15 Q13.16 Q13.17 Q13.18

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