Video: Determining the Variation of Period of an Oscillating Object with a Driving Force Applied

A diver on a diving board is undergoing SHM. Her mass is 55.0 kg and the period of her motion is 0.800 s. The next diver is a male whose period of simple harmonic oscillation is 1.05 s. What is his mass if the mass of the board is negligible?

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Video Transcript

A diver on a diving board is undergoing simple harmonic motion. Her mass is 55.0 kilograms, and the period of her motion is 0.800 seconds. The next diver is a male whose period of simple harmonic oscillation is 1.05 seconds. What is his mass if the mass of the board is negligible?

In this statement, we’re told the mass of the first diver β€” we’ll call that π‘š sub one β€” is 55.0 kilograms. And that this first diver has a period of motion β€” we’ll call that capital 𝑇 sub one β€” of 0.800 seconds.

The next diver in line has a period of simple harmonic oscillation of 1.05 seconds; we’ll call that 𝑇 sub two. We want to know what is the mass of second diver if the mass of the board is negligible; we’ll call that mass π‘š sub two. To begin solving this problem, let’s recall a relationship between the period 𝑇 and mass π‘š involved in a simple harmonic system.

For a simple harmonic oscillator, the period 𝑇 is equal to two times πœ‹ times the square root of the mass on the oscillator divided by the oscillator’s spring constant, π‘˜.

When we consider our first diver, diver one, that diver’s period is equal to two πœ‹ times the square root of the diver’s mass divided by the spring constant π‘˜. We know both 𝑇 one and π‘š one, but we don’t know the spring constant π‘˜.

We’ll want to know that value in order to be able to solve for the mass of the second diver π‘š sub two. So let’s rearrange this equation to solve for π‘˜, the spring constant.

If we square both sides and then multiply both sides of the equation by π‘˜ divided by 𝑇 sub one squared, then the term in parentheses becomes four πœ‹ squared times π‘š one divided by π‘˜, and the power two goes away.

Then on the right-hand side of our equation, the spring constant π‘˜ cancels; and on the left-hand side, the period 𝑑 sub one squared cancels, leaving us with a simplified version of the equation which says π‘˜ is equal to four πœ‹ squared times π‘š sub one divided by 𝑇 sub one squared.

Since π‘š sub one and 𝑇 sub one are both given in our problem statement, we can insert those values into this relationship now. The spring constant π‘˜ is equal to four πœ‹ squared times 55.0 kilograms divided by 0.800 seconds squared. This value is equal to 3393 newtons per meter.

And we’ll keep four significant figures for this value because this is an intermediate step. Now that we know π‘˜, we can move ahead towards solving for π‘š two, the mass of the second diver.

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œ‹ times the square root of diver two’s mass divided by π‘˜.

We want to rearrange this equation to solve for π‘š sub two, the diver’s mass. Let’s do that by again squaring both sides of the equation. On the right side of our equation, this creates an expression which reads four πœ‹ squared times π‘š two divided by π‘˜.

If we then multiply both sides of the equation by π‘˜ divided by four πœ‹ squared, then both instances of four πœ‹ squared and π‘˜ on the right-side of our equation cancel out. We’re left with an expression for π‘š two that reads π‘š two, the mass of the second diver, is equal to π‘˜ divided by four πœ‹ squared multiplied by 𝑇 sub two squared.

We were given 𝑇 sub two as 1.05 seconds and we’ve solved for π‘˜, the spring constant. Let’s insert those values into the equation now. So π‘š sub two, the mass of the second diver, is equal to 3393 newtons per meter divided by four πœ‹ squared multiplied by 1.05 seconds squared.

This mass is equal to 94.7 kilograms. This is the mass of the second diver.

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