Question

A 90.0-m-long brass rod is struck at one end. A person at the other end hears two sounds as a result of two longitudinal waves, one traveling in the metal rod and the other traveling in the air. What is the time interval between the two sounds? Take the speed of sound in air to be 344 m/s. Use 8600 kg/m3 for the density of brass and 9.00×109 Pa for Young's modulus of brass.

Answer 1

Answer:

0.174413400763 s

Explanation:

L = Length of rod = 90 m

Speed of sound in air = 344 m/s

$\rho$ = Density of brass = 8600 kg/m³

Y = Young's modulus = $9\times 10^9\ Pa$

Velocity of sound in a material is given by

$v_b=\sqrt{\dfrac{Y}{\rho}}\\\Rightarrow v_b=\sqrt{\dfrac{9\times 10^9}{8600}}\\\Rightarrow v_b=1022.99150921\ m/s$

Time taken in brass

$t=\dfrac{L}{v_b}\\\Rightarrow t=\dfrac{90}{1022.99150921}\\\Rightarrow t=0.0879772697913\ s$

In air

$t=\dfrac{L}{v}\\\Rightarrow t=\dfrac{90}{343}\\\Rightarrow t=0.262390670554\ s$

The time interval is $0.262390670554-0.0879772697913=0.174413400763\ s$

Answer 2

This question involves the concepts of density, the velocity of sound, and time interval.

The time interval between the two sounds is "0.17 s".

First, we will calculate the time taken by the sound in air:

$t_a=\frac{s}{v_a}$

where,

$t_a$ = time taken by the sound in air = ?

s = distance travelled = 90 m

$v_a$ = velocity of sound in air = 344 m/s

Therefore,

$t_a=\frac{90\ m}{344\ m/s}\\\\t_a = 0.26\ s$

Now, we will calculate the velocity of sound in the brass rod by using the following formula:

$v_b=\sqrt{\frac{E}{\rho}}$

where,

$v_b$ = velocity of sound in brass = ?

E = young's modulus of brass = 9 x 10⁹ Pa

$\rho$ = density of brass = 8600 kg/m³

Therefore,

$v_b=\sqrt{\frac{9\ x\ 10^9\ Pa}{8600\ kg/m^3}}\\\\v_b=1022.99\ m/s$

Now, we will calculate the time taken by the sound in the brass:

$t_b=\frac{s}{v_b}$

where,

$t_b$ = time taken by the sound in the brass rod = ?

s = distance travelled = 90 m

$v_b$ = velocity of sound in the brass rod = 1022.99 m/s

Therefore,

$t_b=\frac{90\ m}{1022.99\ m/s}\\\\t_b = 0.09\ s$

Hence, the time interval will be:

$\Delta t = t_a-t_b$

Δt = 0.26 s - 0.09 s

Δt = 0.17 s

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults