A metal rod of length 1m is clamped at two points as shown in the figure. Distance of the clamp from the two ends are 5 cm and 15 cm, respectively. Find the minimum and next higher frequency of natural longitudinal oscillation of the rod. Given that Young's modulus of elasticity and density of aluminium are `Y= 1.6 xx 10^(11) Nm^(-2)` and `rho = 2500 kgm^-3` , respectively.

Answer

A, B, D

Explanation

Speed of longitudinal waves in the rod,
`v =sqrt(Y/rho) =sqrt(1.6 xx (10^11))/ (2500)) = 8000 m//s`
At the clamped position nodes will be formed.
Between the clamps integer number of loops will
be formed. Hence,
`n_1 lambda/2 = 80`
or ` n_1lambda = 160`.......(i)

Between P and R, P is a fixed end and R is the free
end. It means the number of loops between P and
R will be odd multiple of `lambda/4` . Then,
`(2n_2 -1)/2 lambda/2 = 5 `
or ` (2n_2 -1)lambda = 20` ...... (ii)
Also between Q and S, lt brgt ` (2n_3 - 1) lambda = 60` ..........(iii)
From Eqs. (i) and (ii), we get
`n_1/ (2n_2 -1) = 160/20 = 8` ........ (iv)
and from Eqs. (i) and (iii).
`(n_1)/(2n_3 - 1) = 160/60 = 8/3` .......(v)
For minimum frequency `n_1 n_2 and n_3` should be
least from Eqs. (iv) and (v).
We get, `n_1 = 8, n_2 = 1, n_3 = 2`
`lambda = 20/(2n_2 -1) = 20 cm ` [from Eq. (ii)]
= 0.2 m
`:. f_(min) = v/lambda = 8000/0.2`
= 40 kHz.
Next higher frequency corresponds to
`n_1 = 24, n_2 = 2 `
and `n_3 = 5 `
`f= 120 kHz`