Derive a relation between time period anf frequency of a wave

Physics

1 answer:

Elza [17]3 years ago

8 0

Answer:

The relation between frequency and time period is given by:

f = 1/T

Explanation:

In a wave motion, the particle move about the mean position with the passage of time. The particles rise to reach the highest point which is crest, and similarly falls to reach the lowest point which is trough. The cycle keeps on repeating.

The time period of the wave can be defined as the time taken to complete one such cycle. Time period is given by:

T = 2π/ω

Frequency can be defined as the number of cycles completed in unit time, which can be taken as the inverse of time period. frequency is given by

f = ω/2π

f = 1/T

You might be interested in

Why did the potassium permanganate crystals start to dissolve in water without being stirred or shaken

KonstantinChe [14]

because the substance in the potassium permanganate crystals are permeable to water, so that means it will dissolve instantly while poured into water

7 0

3 years ago

Read 2 more answers

A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. The correct form is: A $+10\,\mu C$ charge and a $-10\,\mu C$ at a distance of 0.3 meters.

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is: