Answer:
[tex]\textsf{1)\quad 48\;inches}[/tex]
[tex]\textsf{2)\quad$2\sqrt[3]{3}$\; inches}[/tex]
Step-by-step explanation:
Question 1
Define the variables:
- Let x be the length of the string.
- Let y be the rate of vibration of a string under constant tension.
The rate of vibration of a string under constant tension varies inversely with the length of the string:
[tex]\boxed{y \propto \dfrac{1}{x} \implies y=\dfrac{k}{x}\quad\text{for a constant $k$}}[/tex]
Given values:
- x = 24 inches
- y = 128 times per second
Substitute the given values of x and y into the formula and solve for k:
[tex]\implies 128=\dfrac{k}{24}[/tex]
[tex]\implies k=128 \cdot 24[/tex]
[tex]\implies k=3072[/tex]
Therefore, the equation is:
[tex]y=\dfrac{3072}{x}[/tex]
To find the length of a string that vibrates 64 times per second, substitute y = 64 into the equation and solve for x:
[tex]\implies 64=\dfrac{3072}{x}[/tex]
[tex]\implies x=\dfrac{3072}{64}[/tex]
[tex]\implies x=48[/tex]
Therefore, the length of a string that vibrates 64 times per second is 48 inches.
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Question 2
Define the variables:
- Let y be the horsepower (hp) that a shaft can safely transmit.
- Let v be the speed of the shaft (in rpm).
- Let d be the diameter of the shaft (in inches).
The horsepower that a shaft can safely transmit varies jointly with its speed and the cube of the diameter:
[tex]\boxed{y \propto vd^3 \implies y=kvd^3\quad\text{for a constant $k$}}[/tex]
Given values:
- y = 45 hp
- v = 100 rpm
- d = 3 inches
Substitute the given values of y, v and d into the formula and solve for k:
[tex]\implies 45=k \cdot 100 \cdot 3^3[/tex]
[tex]\implies 45=k \cdot 100 \cdot 27[/tex]
[tex]\implies 45=2700k[/tex]
[tex]\implies k=\dfrac{45}{2700}=\dfrac{1}{60}[/tex]
Therefore, the equation is:
[tex]y=\dfrac{vd^3}{60}[/tex]
To find the diameter of the shaft in order to transmit 60 hp at 150 rpm, substitute y = 60 and v = 150 into the equation and solve for d:
[tex]\implies 60=\dfrac{150d^3}{60}[/tex]
[tex]\implies 3600=150d^3[/tex]
[tex]\implies d^3=\dfrac{3600}{150}[/tex]
[tex]\implies d^3=24[/tex]
[tex]\implies d=\sqrt[3]{24}[/tex]
[tex]\implies d=\sqrt[3]{8 \cdot 3}[/tex]
[tex]\implies d=\sqrt[3]{8} \sqrt[3]{3}[/tex]
[tex]\implies d=\sqrt[3]{2^3} \cdot \sqrt[3]{3}[/tex]
[tex]\implies d=2\sqrt[3]{3}[/tex]
Therefore, the diameter of a shaft that transmits 60 hp at 150 rpm is 2³√3 inches.
