A tuba creates a 4th harmonic of frequency 116.5 Hz. When the first

valve is pushed, it opens an extra bit

of tubing 0.721 m long. What is the

new frequency of the 4th harmonic?

(Hint Find the original length.)

(Speed of sound = 343 m/s)

(Unit = Hz)

93.54 Hz

Explanation:

✓From the question, Number of harmonic frequency is 4

✓ the frequency (f₄ )= 116.5 Hz

✓harmonic frequency can be calculated using below expresion

fₙ = [ (nv)/4L]eqn(1)

v = speed of sound= 343 m/s

n = number of given harmonic frequency

L = Length of the rope

Using above expresion ,and substitute the values at (n=4) which is 4th harmonic frequency to find the " initial Lenght of the rope

fₙ = [ (nv)/4L]

f₄ = 4× 343 /4L

f₄ = 343 /L

L= 343 /f₄

But f₄= 116.5 Hz

L= 343/116.5= 2.944m

Hence, initial Lenght of the rope= 2.944m

We can determine the frequency of new length as ( initial Lenght of the rope + tubing Lenght)

= ( 2.944m + 0.721m )

= 3.667m

Hence, new length= 3.667m

To find the new frequency of the 4th harmonic we will use eqn(2)

f₄ = v/l eqn(2)

From equation (2) If we substitute the values we have

f₄ = (343/3.667)

= 93.54 Hz

Hence, the the new frequency of the 4th harmonic is

93.54 Hz

i need the answer to this too but idk it ; -;

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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