Question

The frequency of a tuning fork is 500Hz. Calculate the time period​

Answer 1

Given :

The frequency of a tuning fork is 500Hz.

To find :

The time period.

Solution :

Time taken by a wave to pass through one point is called time period.

The time period and frequency are related as

$\boxed{ \gray{ \bf T = \dfrac{1}{ \upsilon}}}$

here,

• T denotes time period.
• v denotes frequency.

now substituting all the given values,

$\leadsto{ \sf T = \dfrac{1}{ \upsilon}}$

$\leadsto{ \sf T = \dfrac{1}{ 500}}$

$\leadsto \underline{\boxed{ \sf T = 0.002 \: sec}}$

thus, the time period is 0.002 sec.

More to know

The number of waves that pass through a given point in one second is called frequency.

• SI unit of frequency are cycle per second (cps) or Hertz (Hz).

The frequency and wavelength are related as,

• v = C/λ

where, v denotes frequency, λ denotes wavelength and C denotes velocity of light

Answer 2

Answer :-

Given :-

• Frequency = 500 Hz

To Find

• Time period

Solution :-

➩ T = 1/n

➩ T = 1/500

➩ T = 0.002 s

Time period = 0.002 s

Additional information :-

Frequency - The number of vibrations completed by a particle in one second is called its frequency.

Time period - Time taken by a vibrating particle to make one complete vibration is called its time period.

Wavelength - It is distance travelled by a wave during one complete vibration of the vibrating particle.

Amplitude - The maximum displacement of the vibrating particles from its mean position is known as amplitude.