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A diver on a diving board is undergoing simple harmonic motion.
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Her mass is 55.0 kilograms, and the period of her motion is 0.800 seconds.
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The next diver is a male whose period of simple harmonic oscillation is 1.05 seconds.
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What is his mass if the mass of the board is negligible?
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In this statement, weβre told the mass of the first diver β weβll call that π sub one β is 55.0 kilograms.
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And that this first diver has a period of motion β weβll call that capital π sub one β of 0.800 seconds.
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The next diver in line has a period of simple harmonic oscillation of 1.05 seconds; weβll call that π sub two.
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We want to know what is the mass of second diver if the mass of the board is negligible; weβll call that mass π sub two.
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To begin solving this problem, letβs recall a relationship between the period π and mass π involved in a simple harmonic system.
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For a simple harmonic oscillator, the period π is equal to two times π times the square root of the mass on the oscillator divided by the oscillatorβs spring constant, π.
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When we consider our first diver, diver one, that diverβs period is equal to two π times the square root of the diverβs mass divided by the spring constant π.
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We know both π one and π one, but we donβt know the spring constant π.
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Weβll want to know that value in order to be able to solve for the mass of the second diver π sub two.
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So letβs rearrange this equation to solve for π, the spring constant.
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If we square both sides and then multiply both sides of the equation by π divided by π sub one squared, then the term in parentheses becomes four π squared times π one divided by π, and the power two goes away.
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Then on the right-hand side of our equation, the spring constant π cancels; and on the left-hand side, the period π‘ sub one squared cancels, leaving us with a simplified version of the equation which says π is equal to four π squared times π sub one divided by π sub one squared.
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Since π sub one and π sub one are both given in our problem statement, we can insert those values into this relationship now.
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The spring constant π is equal to four π squared times 55.0 kilograms divided by 0.800 seconds squared.
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This value is equal to 3393 newtons per meter.
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And weβll keep four significant figures for this value because this is an intermediate step.
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Now that we know π, we can move ahead towards solving for π two, the mass of the second diver.
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Using the equation for the simple harmonic motion period once again, this time in the case of diver number two, we see that the period for diver two equals two π times the square root of diver twoβs mass divided by π.
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We want to rearrange this equation to solve for π sub two, the diverβs mass.
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Letβs do that by again squaring both sides of the equation.
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On the right side of our equation, this creates an expression which reads four π squared times π two divided by π.
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If we then multiply both sides of the equation by π divided by four π squared, then both instances of four π squared and π on the right-side of our equation cancel out.
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Weβre left with an expression for π two that reads π two, the mass of the second diver, is equal to π divided by four π squared multiplied by π sub two squared.
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We were given π sub two as 1.05 seconds and weβve solved for π, the spring constant.
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Letβs insert those values into the equation now.
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So π sub two, the mass of the second diver, is equal to 3393 newtons per meter divided by four π squared multiplied by 1.05 seconds squared.
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This mass is equal to 94.7 kilograms.
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This is the mass of the second diver.