Question

A metal bar with a length of $1.50\;{\text{m}}$ has density of $6400\;{\text{kg/}}{{\text{m}}^{\text{2}}}$. Longitudinal sound waves take $3.90 \times {10^{ - 4}}\;{\text{s}}$ to travel from one end of the bar to the other. The young’s modulus for this metal is $94.7 \times {10^x}\;{\text{Pa}}$ . Find the value of $x$.

Answer 1

The young’s modulus of the metal is proportional to the velocity of the longitudinal sound waves through the metal. Also the velocity can be determined from the distance the wave travelled and time taken to travel. These two equations can be equated to find the young’s modulus of the metal.

Complete step-by-step solution :
Given the density of the metal $\rho = 6400\;{\text{kg/}}{{\text{m}}^{\text{2}}}$ . The length of the bar is $l = 1.50\;{\text{m}}$ .
The time taken for the longitudinal sound waves to travel from one end of the bar to the other is $t = 3.90 \times {10^{ - 4}}\;{\text{s}}$
Therefore the speed of the longitudinal wave is given as,
$v = \dfrac{l}{t}$
Where, $l$ is the distance traveled and it is equal to the length of the metal rod and $t$ is the time taken for the longitudinal sound waves to travel from one end of the bar to the other.
Substitute the values in the above expression.
$\ v = \dfrac{{1.50\;{\text{m}}}}{{3.90 \times {{10}^{ - 4}}\;{\text{s}}}} \\\ = 3.846 \times {10^3}\;{\text{m/s}} \\\$
Also the speed of the longitudinal wave is given as,
$v = \sqrt {\dfrac{Y}{\rho }}$
Where, $Y$ is the young’s modulus and $\rho$ is the density.
Substitute the values for velocity and density in the above expression.
$\ 3.846 \times {10^3}\;{\text{m/s}} = \sqrt {\dfrac{Y}{{6400\;{\text{kg/}}{{\text{m}}^{\text{2}}}}}} \\\ {\left( {3.846 \times {{10}^3}\;{\text{m/s}}} \right)^2} = \dfrac{Y}{{6400\;{\text{kg/}}{{\text{m}}^{\text{2}}}}} \\\ Y = 9.466 \times {10^{10}}\;{\text{Pa}} \\\ {\text{ = 94}}{\text{.7}} \times {10^9}\;{\text{Pa}} \\\$
Therefore the value of $x$ is $9$ .

Note:- If the density of the metal is greater, the velocity of the longitudinal wave will be less. Thus the metal having less density will be good for the sound waves to travel.