Question

A tuning fork makes 100vibrations in 5 sec. Find
its frequency and time period​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Vibration of tuning fork = 100 in 5 sec

$\color{darkblue}\underline{\underline{\sf To\:Find-}}$

• Frequency
• Time Period

➩ Frequency (f)

It is defined as number of wave formed per second.

$\Large\color{indigo}\bullet\boxed{\sf f=\dfrac{vibration}{time}}$

$\implies{\sf f=\dfrac{100}{5} }$

$\color{red}\implies{\sf f= 20Hz}$

➩ Time Period (T)

Defined as number of seconds per vibration

i.e

$\Large\color{indigo}\bullet\boxed{\sf T=\dfrac{1}{f}}$

$\implies{\sf T=\dfrac{1}{20}}$

$\color{red}\implies{\sf T=0.05sec}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

$\bullet$ Frequency (f) = 20Hz

$\bullet$ Time period (T) = 0.05sec

Answer 2

Answer:

In this case a tuning fork vibrates 100 times in 5 sec.

Explanation:

the frequency is given as f= 100/5 = 20 Hz

as we know, time period is reciprocal of frequency. Hence T= 1/F

=1/20 = 0.5s

Hence the frequency is 20Hz and the time period is 0.5s.