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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

Arun
25763 Points
11 months 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
14149 Points
11 months 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