Explanation:
It is given that,
Frequency of the tuning fork, f₁ = 200 Hz
Beat frequency, [tex]f_b=4\ Hz[/tex]
The difference of frequencies is called the beat frequency of the tuning fork. It is given by :
[tex]|f_1-f_2|=4[/tex]
[tex]200-f_2=\pm 4[/tex]
[tex]f_2=196\ Hz[/tex]
or
[tex]f_2=200\ Hz[/tex]
So, the correct option is (b) "196 hertz". Hence, this is the required solution.