Verified Solution • 2025–26 Session moderate

Q13.16: NCERT Physics - Part 2 OSCILLATIONS 13.16 Solution: A simple pendulum of length l and having a bob of mass M is…

A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Step-by-Step Verified Solution

**Answer:** T=2πlg2+(v2/R)2T = 2\pi\sqrt{\frac{l}{\sqrt{g^2 + (v^2/R)^2}}} **Step-by-Step Explanation:** The bob experiences gravity g downward and centripetal acceleration v2/Rv^2/R horizontally. Effective acceleration:
aeff=g2+(v2R)2a_{eff} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}
Time period:
T=2πlaeff=2πlg2+(v2/R)2T = 2\pi\sqrt{\frac{l}{a_{eff}}} = 2\pi\sqrt{\frac{l}{\sqrt{g^2 + (v^2/R)^2}}}
[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

Related Learning Resources & Exam Preparation

Chapter Practice Bank
Practice all topic-wise questions from OSCILLATIONS
Reference Textbook Solutions
Explore complete chapter exercises for NCERT Physics - Part 2
Competitive PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring