If pendulum A makes 15 oscillations in 5 seconds, and pendulum B makes 12 oscillations in 3 seconds. Which pendulum has a higher frequency  A or B?

Here, we will calculate the frequencies of pendulum A and pendulum B.
The frequency of pendulum A (f_A) is given by,
     $\mathrm{Frequency}\left({\mathrm{f}}_{\mathrm{A}}\right)=\frac{\mathrm{no}. \mathrm{of} \mathrm{oscillations}}{\mathrm{time} \mathrm{taken}}$   

 [where $\mathrm{n}=\mathrm{no}. \mathrm{of} \mathrm{oscillations}$, $\mathrm{t}=\mathrm{time} \mathrm{taken}$]
$\Rightarrow \mathrm{Frequency} \left({\mathrm{f}}_{\mathrm{A}}\right)=\frac{15}{5}$ 

$\Rightarrow \mathrm{Frequency} \left({\mathrm{f}}_{\mathrm{A}}\right)=3\mathrm{Hz}$     …………………….. (i)
Now, the frequency of pendulum B (f_B) is given by
     $\mathrm{Frequency} \left({\mathrm{f}}_{\mathrm{B}}\right)=\frac{\mathrm{no}. \mathrm{of} \mathrm{oscillations}}{\mathrm{time} \mathrm{taken}}$
$\Rightarrow \mathrm{Frequency} \left({\mathrm{f}}_{\mathrm{B}}\right)=\frac{12}{3}$
$\Rightarrow \mathrm{Frequency} \left({\mathrm{f}}_{\mathrm{B}}\right)=4 \mathrm{Hz}$     …………….......………… (ii)
From equations (i) and (ii), we can say (f_B) > (f_A)
Therefore, pendulum B has a higher frequency as compared to pendulum A.