Question

An electric vibrator produces ripples in a ripple tank,such that distance between one crest and one trough is 4 cm if the vibrations are produced at a rate of 4800 per minute calculate the time period and wave velocity

Answer 1

Given:

An electric vibrator produces ripples in a ripple tank,such that distance between one crest and one trough is 4 cm if the vibrations are produced at a rate of 4800 per minute.

To find:

Time period and wave velocity ?

Calculation:

We know that time period refers to the time taken to complete one full oscillation by a particle in a wave.

$\rm \therefore \: time \: period = \dfrac{1}{frequency}$

$\rm \implies \: time \: period = \dfrac{1}{( \frac{4800}{60}) }$

$\rm \implies \: time \: period = \dfrac{1}{80 }$

$\rm \implies \: time \: period = 0.0125 \: sec$

So, time period is 0.0125 sec.

Now, let wave velocity be v :

$\rm \therefore \: v = frequency \times wavelength$

$\rm \implies \: v = frequency \times (2 \times distance \: between \: crest \: and \: trough)$

$\rm \implies \: v = 0.0125 \times (2 \times \dfrac{4}{100} )$

$\rm \implies \: v = {10}^{ - 3} \: m {s}^{ - 1}$

So, velocity of wave is 0.001 m/s.

Answer 2

Answer:An electric vibrator produces ripples in a ripple tank,such that distance between one crest and one trough is 4 cm if the vibrations are produced at a rate of 4800 per minute.

To find:

Time period and wave velocity ?

Calculation:

We know that time period refers to the time taken to complete one full oscillation by a particle in a wave.

So, time period is 0.0125 sec.

Now, let wave velocity be v :

So, velocity of wave is 0.001 m/s.

Explanation: