Question

A string vibrates with a frequency of 250Hz. If the distance between two consecutive troughs of a transverse wave produced in the string is 10cm, find the velocity and time period of the wave.

Answer 1

Answer:

given wavelength is 10 cm

time period 1/ f= 4× 10 raised minus 4

speed is f lamda= 10×250= 2500 cm per sec

Answer 2

Given:

A string vibrates with a frequency of 250Hz. The distance between two consecutive troughs of a transverse wave produced in the string is 10cm.

To find:

Velocity and time period of the wave.

Calculation:

Time period of a wave is defined as the time taken by the oscillatory particle (of the wave) to complete one full oscillation. It is the reciprocal of frequency.

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

$= > \rm{time \: period = \dfrac{1}{ 250} }$

$= > \rm{time \: period = \dfrac{1000}{250} \times {10}^{ - 3} }$

$\boxed{ = > \rm{time \: period = 4 \times {10}^{ - 3} \: sec}}$

Now , distance between consecutive trough represents the wavelength = 10 cm.

$\rm{velocity = \lambda \times frequency}$

$= > \rm{velocity = \dfrac{10}{100} \times 250}$

$\boxed{ = > \rm{velocity = 25 \: m {s}^{ - 1} }}$

Hope It Helps.