A tuning fork of frequency 480 Hz produces 10 beats per second when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces lesser beats per second than before

Options

A. 460 Hz
B. 470Hz
C. 480 Hz
D. 490 Hz

Answer

B

Explanation

If suppose `n_(s)` = frequency of string = `(1)/(2l) sqrt((T)/(m))`
`n_(f)` = Frequency of turning fork = 480 Hz
x = Beats heard per second = 10
as tension T increases, so `n_(s)` increase(uparrow)
Also it is given that number of beats per sec decreases (i.e, x downarrow)
Hence `n_(s)uparrow - n_(f) = x downarrow ....(i) rightarrow` wrong
`n_(f) - n_(s)uparrow = x downarrow ....(ii) rightarrow` Correct
implies ` n_(s) = n_(f) - x = 480 - 10` = 470 Hz.