AIPMT MAINS2009Physics-Waves
AIPMT MAINS 2009 Physics Beats MCQ Question
Type: MCQ-numerical-Medium-Class 11
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is:
A
7
B
8
C
3
D
5
Correct Answer
Option A
Detailed Explanation
The beat frequency is the difference in frequencies of the two strings. Using the formula for frequency of a string, f = (1/2L)√(T/μ), where L is the length, T is the tension, and μ is the mass per unit length, we calculate the frequencies and find their difference to be 7 Hz.
Found an issue with this question?