The total length of a sonometer wire fixed between two bridges is 110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6:3:2. If the tension in the wire is 400 N and the mass per unit length of the wire is 0.01 kg m -1 , then the minimum common frequency with which all the three parts can vibrate, is
Text Solution
Verified by ExpertsA
1000 Hz
The frequency in any 
mode for a segment is 
constant no of loops 
length of the segment so, no of loops are in the ratio 6:3:2 hence total loops 
The string is divided in 
part such that for minimum frequency, the wavelength is maximum
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems