Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
364 views
in Laws of motion by (15.2k points)
closed by

A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits −φ and + φ . For an angular displacement θ (|θ| <φ) , the tension in the string and the velocity of the bob are T and v respectively. The following relations hold good under the above conditions.

(A) T cos\(\theta\) = Mg

(B) T - mg cos\(\theta\) = \(\frac{Mv^2}{L}\)

(C) The magnitude of the tangential acceleration of the bob |\(\alpha_\tau\)| = g sin\(\theta\)

(D) T = Mg cos\(\theta\)

2 Answers

+1 vote
by (39.0k points)
selected by
 
Best answer

Correct options are (B) and (C)

Motion of pendulum is part of a circular motion. In circular motion it is better to resolve the forces in two perpendicular directions. First along radius (towards centre) and second along tangential. Along radius net force should be equal to \(\frac{mv^2}R\) and along tangent it should be equal to maT, where aT is the tangential acceleration in the figure.

T – Mg cos θ = \(\frac{mv^2}L\) and

Mg sin θ = MaT

or aT = g sinθ

+1 vote
by (15.7k points)

Answer is :

 (B) T - mg cos\(\theta\) = \(\frac{Mv^2}{L}\)

 (C) The magnitude of the tangential acceleration of the bob |\(\alpha_\tau\)| = g sin\(\theta\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...