Question

1. A wave is introduced into a thin wire held tight at each end. It has amplitude of 3.8 cm, a
frequency of 51.2 Hz and a distance from a crest to the neighbouring trough of 12.8 cm.
Determine the period of such a wave.​

Answer 1

Answer:

See below.

Explanation:

We are given the frequency of the wave,

$f=51.2Hz$

The period is given by,

$T=\dfrac {1}{f}=\dfrac {1}{51.2}=0.0195s$

Answer 2

Answer:

The time period of such a wave is $T=0.0195 \mathrm{~s}$

Explanation:

A sinusoidal wave's frequency is defined as the number of complete oscillations made by any wave constituent per unit of time.

We may comprehend that if a body is in periodic motion, it has completed one cycle after passing through a series of events or locations and returning to its original condition using the notion of frequency.

As a result, frequency is a term that describes the pace at which oscillations and vibrations occur.

The frequency is given by $5.12Hz$

The following is the relationship between period and frequency:

$T=\frac{1}{f}$                $[f=\text { fiequency }]$

Now substitute the value $52.1Hz$ in above equation .

$\begin{aligned}T &=\frac{1}{51.2 \mathrm{~Hz}} \\&=0.0195 \mathrm{~s}\end{aligned}$