How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenly tempered chromatic scale) on a day when air temperature is 20.0°C ? It is open at both ends.

L=0.654 m

Explanation:

Concepts and Principles  

1- The speed of sound in air is expressed as a function of the temperature of air as follows:  

 v=(331 m/s)√(1+T_C/273°C)                        (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.  

Standing Wave Patterns in Pipes:  

A pipe open at both ends can have standing wave patterns with resonant frequencies:  

f=v/λ=nv/2L                     n=1,2,3.........

where v is the speed of sound in air.  

Given Data

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C  

The flute is open at both ends.  

Required Data

We are asked to determine the length of the tube.  

Solution  

The speed of sound in air at temperature T = 20°C is found from Equation (1):

 v=(331 m/s)√(1+T_C/273°C)  

 =342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):  

f=v/2L

Solve for L:  

L=v/2f_1

L=0.654 m

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.