Question

Find the fundamental frequency and next three frequencies that could cause a standing wave pattern on a string that is ${{30}}{{.0m}}$ long has a mass per unit length of ${{9}}{{.00 \times 1}}{{{0}}^{{{ - 3}}}}{{kg/m}}$ and is stretched to a tension of ${{20N}}{{.}}$

Answer 1

Hint: When two progressive wave of same amplitude, same wavelength and same frequencies travelling with same speed along same straight line in opposite direction superimpose, a new set of waves are formed called stationary waves and there is a particular minimum frequency for which this standing wave formed called as fundamental frequency which depends upon length, tension and mass per unit length of string.

Formula used:
(i) ${{{\upsilon }}_{{1}}}\;{{ = }}\;\dfrac{{{1}}}{{{{2L}}}}\sqrt {\dfrac{{{T}}}{{{m}}}{{,}}} $ where ${{{\upsilon }}_{{1}}}$ represents fundamental frequency,
${{'L'}}$ is length of string,
${{'T'}}$ is tension in string
${{'m'}}$ is mass per unit length of string
(ii) ${{{\upsilon }}_{{2}}}{{ = 2}}{{{\upsilon }}_{{1}}}$ Where ${{{\upsilon }}_{{2}}}$ is next frequency than ${{{\upsilon }}_{{1}}}$ which can cause standing wave pattern
    ${{{\upsilon }}_{{3}}}{{ = 3}}{{{\upsilon }}_{{1}}}$ Where ${{{\upsilon }}_{{3}}}$ is next frequency than ${{{\upsilon }}_{{2}}}$ which can produce standing wave pattern.

Complete step by step solution:
A standing wave, also known as a stationary wave, is a wave which oscillates in time but its peak amplitude value does not move in space. The peak amplitude of the wave oscillations at any point in space is always constant with time, and the oscillations at different points are in phase throughout the motion. The velocity of a traveling wave for a string when it is stretched is determined by the tension and the mass per unit length of the string.

We know, we have given length of string ${{L}}\;{{ = }}\;{{30m}}$
Tension in the string ${{ = }}\;{{20N}}$
And mass per unit length ${{ = }}\;{{9 \times 1}}{{{0}}^{{{ - 3}}}}{{kg/m}}$
So, by the relation of fundamental frequency ${{{\upsilon }}_{{1}}}$ , we know,
${{{\upsilon }}_{{1}}}\;{{ = }}\;\dfrac{{{1}}}{{{{2L}}}}\sqrt {\dfrac{{{T}}}{{{m}}}} $
So, by substituting the values of ${{L,}}\;{{m}}$ and ${{T}}$ respectively we get,
${{{\upsilon }}_{{{1}}\;}}{{ = }}\;\dfrac{{{1}}}{{{{2 \times 30}}}}\sqrt {\dfrac{{{{20}}}}{{{{9 \times 1}}{{{0}}^{{{ - 3}}}}}}} \;{{ = }}\;{{0}}{{.786}}{{{H}}_{{z}}}$
${{{n}}^{{{th}}}}$ order harmonics is given by ${{{\upsilon }}_{{n}}}\;{{ = }}\;{{n}}{{{\upsilon }}_{{1}}}$
So, for ${{n}}\;{{,1}}\;{{,2}}\;{{,3}}\;{{,4}}\,$ respective frequencies are
${{{\upsilon }}_{{2}}}\;{{ = }}\;{{1}}{{.57}}{{{H}}_{{z}}}$
${{{\upsilon }}_{{{3}}\;}}{{ = }}\;{{2}}{{.358}}{{{H}}_{{z}}}$
${{{\upsilon }}_{{{4}}\;}}{{ = }}\;{{3}}{{.144}}{{{H}}_{{z}}}$

Note: If you have given the mass of string only then before substituting into formula first find mass per unit length by dividing it with length of string.