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A stretched wire of 60 cm vibrates with a fundamental note of frequency 256 Hz. If the length of the wire is decreased to 15 cm while the tension in kept constant. The fundamental frequency of the wire will be(a) 64 Hz(b) 256 Hz(c) 512 Hz(d) 1024 Hz

sumit kumar , 10 Years ago
Grade Upto college level
anser 2 Answers
Arun

Last Activity: 4 Years ago

Frequency of fundamental note is given by 
Here, ν is the frequency of fundamental note , v is speed of wave , l is the length of string , T is tension in the string and μ is mass density { mass per unit length}
in first case , 
---(1)
in second case, 
----(2)
From equations (1) and (2) 
256/320 = 2(l-10)/2l 
⇒ 4/5 = ( l - 10)/l 
⇒ 4l = 5( l - 10) 
⇒ 4l = 5l - 50 ⇒ l = 50 cm 
Hence, length of string is 50 cm
 

Vikas TU

Last Activity: 4 Years ago

Dear student 
The above ans is not properly written 
Frequency of fundamental note is given by
Here, ν is the frequency of fundamental note , v is speed of wave , l is the length of string , T is tension in the string and μ is mass density { mass per unit length}
in first case ,
256 = v/21 sqrt(T/mew)---(1)
in second case,
320 = v/2(1-10) sqrt(T/mew)----(2)
From equations (1) and (2)
256/320 = 2(l-10)/2l
⇒ 4/5 = ( l - 10)/l
⇒ 4l = 5( l - 10)
⇒ 4l = 5l - 50 ⇒ l = 50 cm
Hence, length of string is 50 cm

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