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When the tension in a string is increased by 44%,the frequency increased by10Hz the frequency of the string is?

saurabh reddy , 9 Years ago
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anser 3 Answers
Abhishek Kumar

Last Activity: 9 Years ago

frequency = nV/2L or, (2n+1)V/4L
v = \sqrt{T/\mu }
If you increase frequency by 44%,
Velocity becomes = \sqrt{1.44T/\mu } = 1.2v
So, frequency changes by 20 %

Mahita

Last Activity: 7 Years ago

f=nv/2l. ......(v=√T/√mue)f=n√T/2l√muef1/f2=T1/T2f+10/f=√144/√100By solving for f we get the value as 50 HZSo..........f=50HZ

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the solution to your problem.
 
f = nv/2l.
(v = √T/√mue)
f = n√T/2l√mue
f1/f2 = T1/T2
f+10/f = √144/√100
By solving for f we get the value as, f = 50 Hz
So, f = 50 Hz
 
Thanks and regards,
Kushagra
 

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