AIPMT PRELIMS2009Physics-Waves
AIPMT PRELIMS 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 calculated using the difference in frequencies of the two strings. The formula for frequency involves tension and mass per unit length, and the difference gives the beat frequency.
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