Question

PLEASE HELP ME !!!!!

1 ) The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string that vibrates 64 times per second?

48 inches
46 inches
42 inches
55 inches

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2 ) The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute - rpm) and the cube of the diameter. If the shaft of a certain material 3 inches in diameter can transmit 45hp at 100 rpm, what must the diameter be in order to transmit 60hp at 150rpm?

24

`\sqrt[3]{150}`

`2\sqrt[3]{3}`

`4\sqrt[3]{6}`

Answer 1

`\textsf{1)\quad 48\;inches}`

`\textsf{2)\quad$2\sqrt[3]{3}$\; inches}`

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:

`\boxed{y \propto (1)/(x) \implies y=(k)/(x)\quad\text{for a constant $k$}}`

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:

`\implies 128=(k)/(24)`

`\implies k=128 \cdot 24`

`\implies k=3072`

Therefore, the equation is:

`y=(3072)/(x)`

To find the length of a string that vibrates 64 times per second, substitute y = 64 into the equation and solve for x:

`\implies 64=(3072)/(x)`

`\implies x=(3072)/(64)`

`\implies x=48`

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:

`\boxed{y \propto vd^3 \implies y=kvd^3\quad\text{for a constant $k$}}`

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:

`\implies 45=k \cdot 100 \cdot 3^3`

`\implies 45=k \cdot 100 \cdot 27`

`\implies 45=2700k`

`\implies k=(45)/(2700)=(1)/(60)`

Therefore, the equation is:

`y=(vd^3)/(60)`

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:

`\implies 60=(150d^3)/(60)`

`\implies 3600=150d^3`

`\implies d^3=(3600)/(150)`

`\implies d^3=24`

`\implies d=\sqrt[3]{24}`

`\implies d=\sqrt[3]{8 \cdot 3}`

`\implies d=\sqrt[3]{8} \sqrt[3]{3}`

`\implies d=\sqrt[3]{2^3} \cdot \sqrt[3]{3}`

`\implies d=2\sqrt[3]{3}`

Therefore, the diameter of a shaft that transmits 60 hp at 150 rpm is 2³√3 inches.