A sonometer strings `AB` of length ` 1m` is stretched by a load and the tension `T` is adjusted so that the string resonates to a frequency of `1 kHz`. Any point `P` of the wire may be held fixed by use of a movable bridge that can slide along the base of sonometer.

Options

A. If point `P` is fixed so that `AP : PB : 1 : 4`, then the smallest frequency for which the sonometer wire resonates is `5 kHz`.
B. If `P` be taken at midpoint of `AB` and fixed , then when the wire vibrates in the third harmonic of its fundamental , the number of nodes in the wire ( including `A and B`) will be totally seven.
C. If the fixed point `P` divides `AB` in the ratio `1 : 2`, then the tension needed to make the string vibrate at `1 kHz` will be `3 T`. (neglecting the terminal effects)
D. The fundamental frequency of the sonometer wire when `P` divides `AB` in the ratio `a : b` will be the same as the fundamental frequency when `P` divides `AB` in the ratio `b : a`.

Answer

A, B, D

Explanation

If `P` divides `AB` in ratio `1 : 4`, then the fundamental frequency corresponds to `5` loops , one loop in `AP` and ` 4` loops in `PB` which corresponds to ` 5 th` harmonic of `1 kHz`. Hence fundamental ` = 5 kHz`.
If `P` be taken at midpoint , the third harmonic will have three loops in each half of the wire `AB`. Hence total number of nodes (including `A` and `B` ) will be ` 5 + 2 = 7`.
If `P` divides `AB` in the ratio `1 : 2`, the fundamental will have three loops , corresponding to the frequency of ` 3 kHz`. For this string to vibrate with the fundamental of ` 1 kHz` , the tension must be `(T//9)`.
The wire `AB` will be symmetry , vibrates with the same fundamental frequency when `P` divides `AB` in the ratio `a:b` or in the ratio` b:a`.