Question

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 43.0 vibrations in 34.0 s. Also, a given maximum travels 430 cm along the rope in 9.0 s. What is the wavelength?

Answer 1

Concepts and reason

The concepts used in this problem are wavelength and frequency.

For determining the wavelength, first calculate the frequency and speed of wave using number of vibrations and then calculate the wavelength using the relation between wavelength, frequency and speed of wave.

Fundamentals

Wavelength:

The wavelength is the length of the wave which is inversely proportional to the frequency of wave. Mathematically,

$\lambda = \frac{v}{f}$

Here, $f$ is the frequency of wave, $v$ is the speed of wave and $\lambda$ is the wavelength.

Frequency:

The occurrences of an event is being repeated. The number that how many times it has been repeated is the frequency. It is calculated as:

$f = \frac{n}{t}$

Here, $n$is the number of vibrations and $t$is the time taken in vibrations.

The frequency of wave is:

$f = \frac{n}{t}$

Substitute $43$ for $n$and $34{\rm{ s}}$ for $t$.

$\begin{array}{c}\\f = \frac{{43}}{{34{\rm{ s}}}}\\\\ = 1.26{\rm{ }}{{\rm{s}}^{ - 1}}\\\end{array}$

The speed of the wave is:

$v = \frac{d}{{t'}}$

Here, $d$ is the distance travelled by wave and $t'$ is the time taken in travelling the distance.

Substitute $9{\rm{ s}}$ for $t'$ and $430{\rm{ cm}}$ for $d$.

$\begin{array}{c}\\v = \frac{{430{\rm{ cm}}}}{{9{\rm{ s}}}}\\\\ = 47.78{\rm{ cm/s}}\\\end{array}$

The wavelength is:

$\lambda = \frac{v}{f}$

Substitute $47.78{\rm{ m/s}}$ for $v$and $1.26{\rm{ }}{{\rm{s}}^{ - 1}}$ for $f$.

$\begin{array}{c}\\\lambda = \frac{{47.78{\rm{ cm/s}}}}{{1.26{\rm{ }}{{\rm{s}}^{ - 1}}}}\\\\ = {\bf{37}}{\bf{.9 cm}}\\\end{array}$

Ans:

The value corresponding to the wavelength is${\bf{37}}{\bf{.9 cm}}$.