Questions
A wire length L and mass per unit length is put under tension of . Two consecutive frequencies that it resonates at are: and . Then L in meters is
detailed solution
Correct option is B
The difference in the frequencies of two successive harmonics is equal to fundamental frequency n=490−420=70Hzn=12LTμ ⇒70=12×L5406×10−3 ⇒L=1140×300=2.142 mTalk to our academic expert!
Similar Questions
A wire having a linear density of 5 g/m is stretched between two rigid supports with a tension of 450 N. It is observed that the wire resonates at a frequency of 420 Hz. The next highest frequency at which the same wire resonates is 490 Hz. Find the length of the wire. (in m)
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
