Figure shown a clamped metal string of length `30 cm` and linear mass density `0.1 kg //m`. Which is taut at a tension of `40N`. A small rider `(` piece of paper `)` is placed on string at point `P` as shown. An external vibrating tuning fork is brought near this string and oscillations of rider are carefully observed.

Now if the tension in the string is made `160N`, at which of the following frequencies of turning fork , rider will not vibrate at all

Options

A. `(100)/(3)`
B. 50 Hz
C. 200 Hz
D. None of these

Answer

C

Explanation

Wave velocity in string is
`V=sqrt((T)/(mu))=sqrt((40)/(0.1))=20m//s`
Fundamental frequency of string oscillations is
`n_(0)=(v)/(2e)=(20)/(0.6)=(100)/(3)Hz`
Thus string will be in resonance with a turning fork of frequency
`n_(f)=(100)/(3)Hz,(200)/(3)Hz,100Hz,(400)/(3)Hz`....
Here rider will not oscillate at all only if it is at a node of stationary wave in all other cases of resonance and non-resonance it will vibrate at the frequency of tuning fork. At a distance `(l)/(3)` from one end node will appear at `3^(rd),6^(th),9^(th)` similar higher Harmonics i.e. at frequencies 100 Hz,200 Hz
If string is divided in odd no of segments these segments can never resonance simultaneously hence at the location of rider antinode is never obtained at any frequency.