Question

You decide to make one of the world’s largest musical instruments, using long cylindrical steel rods that are clamped tightly at both ends. The speed of longitudinal waves in rods of the type of steel you use is 5068 m/s and the steel has a Young’s modulus of 199 GPa.

Required:
a. What is the density of the steel, in kilograms per cubic meter?
b. When a longitudinal wave is excited in one of the steel rods in your instrument to produce resonance, will the rod’s ends be nodes or antinodes?
c. A standard frequency for tuning musical instruments is 440 Hz for the pitch of A above middle C, denoted A4. What is the length, in meters, of the steel rod that produces the pitch A4 as its fundamental longitudinal resonance?
d. Assume, instead, that you want to create a smaller version of this, using the same kind of construction, with one rod of length 9.5 cm. What would the fundamental longitudinal resonance frequency of that rod be, in hertz, with the same longitudinal wave speed as above, 5149 m/s?

Answer 1

Answer:

Explanation:

speed of longitudinal wave in a rod = $\sqrt{\frac{Y}{D} }$

Where Y is young's modulus of elasticity and D is density .

Putting given data

5068 = $\sqrt{\frac{199\times 10^{9}}{D} }$

D = 199 x 10⁹ / .2568 x 10⁸

= 7749 kg / m³

b )

Rod's ends are not to vibrate as they are clamped . So node is produced there .

c )

Frequency = 440 Hz

Note = fundamental

l = λ / 2  , l is length of rod and λ is wavelength

wavelength of 440 Hz note

= velocity of sound / frequency

= 5068 / 440

= 11.518 m

l = 11.518 / 2

= 5.76 m .

Length of rod = 5.76 m

d )

l = .095 m

velocity = 5149 m /s

wavelength = 2l

= 2 x .095 m

= .19 m

frequency = velocity / wavelength

= 5149 / .19

= 27100 Hz .

Answer 2

In this exercise we have to use the density knowledge to calculate the wave speed, like this:

A)$D=7749 kg/m^3$

B) Have node.

C) $L = 5.76 m$
D) $f= 27100 Hz$

So knowing that from the data informed by the statement we find that:

A) First we have to calculate the density as:

$D = 199 * 10^9 / 0.2568 * 10^8\\D= 7749 kg / m^3$

B) Rod's ends are not to vibrate as they are clamped.

C) Second we have to calculate the length as:

$Length= velocity \ of \ sound / frequency\\= 5068 / 440\\= 11.518 m\\L= 11.518 / 2\\L= 5.76 m$

D) Third  we have to calculate the frequency as:

$wavelength = 2L\\= 2 * 0.095 m\\=0 .19 m\\frequency = velocity / wavelength\\= 5149 / 0.19\\f= 27100 Hz .$

See more about waves at brainly.com/question/3004869