Math / Algebra

QuestionAn air-track glider is attached to a spring. The glider is pulled to the right and released from rest at $t=0 \mathrm{~s}$ . It then oscillates with a period of 240 s and a maximum speed of $60.0 \mathrm{~cm} / \mathrm{s}$
Part A
What is the amplitude of the oscillation? Express your answer with the appropriate units. View Available Hint(s) $\square$ Submit Previous Answers * Incorrect; Try Again; 9 attempts remaining
Part B
What is the glider's position at $t=0.300 \mathrm{~s}$ ? Express your answer with the appropriate units. View Available Hint(s) $\square$ Submit

Studdy Solution

STEP 1

What is this asking? We need to find how far the glider moves from the center and where it is a short time after letting go. Watch out! Remember, the glider starts at its furthest point, not the middle!

STEP 2

1. Find the amplitude
2. Find the position at $t = 0.300 \mathrm{~s}$

STEP 3

Alright, so we're dealing with an oscillating glider.
It's attached to a spring, gets pulled, and whoosh!
It goes back and forth.
We know the period is $T = 240 \mathrm{~s}$ and its maximum speed is $v_{max} = 60.0 \mathrm{~cm/s}$ .
We want to find the amplitude, which is how far it moves from the center point.

STEP 4

The formula that connects these is $v_{max} = \omega \cdot A$ , where $\omega$ is the angular frequency and $A$ is the amplitude.
Think of $\omega$ like how fast the glider is oscillating back and forth.

STEP 5

We can find $\omega$ using the period $T$ .
The formula is $\omega = \frac{2\pi}{T}$ .
Plugging in our period $T = 240 \mathrm{~s}$ , we get $\omega = \frac{2\pi}{240 \mathrm{~s}} = \frac{\pi}{120 \mathrm{~s}}$ So, the glider oscillates $\frac{\pi}{120}$ radians every second.

STEP 6

Now, let's plug this $\omega$ back into our maximum speed formula: $v_{max} = \omega \cdot A$ .
We know $v_{max} = 60.0 \mathrm{~cm/s}$ , so $60.0 \mathrm{~cm/s} = \frac{\pi}{120 \mathrm{~s}} \cdot A$

STEP 7

To solve for $A$ , we multiply both sides by $\frac{120 \mathrm{~s}}{\pi}$ : $A = \frac{60.0 \mathrm{~cm/s} \cdot 120 \mathrm{~s}}{\pi} = \frac{7200 \mathrm{~cm}}{\pi} \approx 2292 \mathrm{~cm}$ So, the amplitude is approximately $2292 \mathrm{~cm}$ !

STEP 8

Now, we want to find the glider's position at $t = 0.300 \mathrm{~s}$ .
Since the glider starts at its maximum position to the right (because it was pulled to the right), we can use the cosine function to describe its motion: $x(t) = A \cos(\omega t)$

STEP 9

We already know $A \approx 2292 \mathrm{~cm}$ and $\omega = \frac{\pi}{120 \mathrm{~s}}$ .
Plugging in $t = 0.300 \mathrm{~s}$ , we get: $x(0.300 \mathrm{~s}) = 2292 \mathrm{~cm} \cdot \cos\left(\frac{\pi}{120 \mathrm{~s}} \cdot 0.300 \mathrm{~s}\right)$

STEP 10

Simplifying inside the cosine: $x(0.300 \mathrm{~s}) = 2292 \mathrm{~cm} \cdot \cos\left(\frac{0.300\pi}{120}\right) = 2292 \mathrm{~cm} \cdot \cos\left(\frac{\pi}{400}\right)$

STEP 11

Calculating the cosine value: $x(0.300 \mathrm{~s}) \approx 2292 \mathrm{~cm} \cdot 0.99996875 \approx 2292 \mathrm{~cm}$ So, at $t = 0.300 \mathrm{~s}$ , the glider is almost at its starting position, approximately $2292 \mathrm{~cm}$ to the right.

STEP 12

Part A: The amplitude of the oscillation is approximately $2292 \mathrm{~cm}$ .
Part B: The glider's position at $t = 0.300 \mathrm{~s}$ is approximately $2292 \mathrm{~cm}$ to the right of the equilibrium position.

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Studdy Solution

STEP 1

What is this asking? We need to find how far the glider moves from the center and where it is a short time after letting go. Watch out! Remember, the glider starts at its furthest point, not the middle!

STEP 2

1. Find the amplitude
2. Find the position at $t = 0.300 \mathrm{~s}$

STEP 3

STEP 4

The formula that connects these is $v_{max} = \omega \cdot A$ , where $\omega$ is the angular frequency and $A$ is the amplitude.
Think of $\omega$ like how fast the glider is oscillating back and forth.

STEP 5

We can find $\omega$ using the period $T$ .
The formula is $\omega = \frac{2\pi}{T}$ .
Plugging in our period $T = 240 \mathrm{~s}$ , we get $\omega = \frac{2\pi}{240 \mathrm{~s}} = \frac{\pi}{120 \mathrm{~s}}$ So, the glider oscillates $\frac{\pi}{120}$ radians every second.

STEP 6

Now, let's plug this $\omega$ back into our maximum speed formula: $v_{max} = \omega \cdot A$ .
We know $v_{max} = 60.0 \mathrm{~cm/s}$ , so $60.0 \mathrm{~cm/s} = \frac{\pi}{120 \mathrm{~s}} \cdot A$

STEP 7

To solve for $A$ , we multiply both sides by $\frac{120 \mathrm{~s}}{\pi}$ : $A = \frac{60.0 \mathrm{~cm/s} \cdot 120 \mathrm{~s}}{\pi} = \frac{7200 \mathrm{~cm}}{\pi} \approx 2292 \mathrm{~cm}$ So, the amplitude is approximately $2292 \mathrm{~cm}$ !

STEP 8

Now, we want to find the glider's position at $t = 0.300 \mathrm{~s}$ .
Since the glider starts at its maximum position to the right (because it was pulled to the right), we can use the cosine function to describe its motion: $x(t) = A \cos(\omega t)$

STEP 9

We already know $A \approx 2292 \mathrm{~cm}$ and $\omega = \frac{\pi}{120 \mathrm{~s}}$ .
Plugging in $t = 0.300 \mathrm{~s}$ , we get: $x(0.300 \mathrm{~s}) = 2292 \mathrm{~cm} \cdot \cos\left(\frac{\pi}{120 \mathrm{~s}} \cdot 0.300 \mathrm{~s}\right)$

STEP 10

Simplifying inside the cosine: $x(0.300 \mathrm{~s}) = 2292 \mathrm{~cm} \cdot \cos\left(\frac{0.300\pi}{120}\right) = 2292 \mathrm{~cm} \cdot \cos\left(\frac{\pi}{400}\right)$

STEP 11

Calculating the cosine value: $x(0.300 \mathrm{~s}) \approx 2292 \mathrm{~cm} \cdot 0.99996875 \approx 2292 \mathrm{~cm}$ So, at $t = 0.300 \mathrm{~s}$ , the glider is almost at its starting position, approximately $2292 \mathrm{~cm}$ to the right.

STEP 12

Part A: The amplitude of the oscillation is approximately $2292 \mathrm{~cm}$ .
Part B: The glider's position at $t = 0.300 \mathrm{~s}$ is approximately $2292 \mathrm{~cm}$ to the right of the equilibrium position.

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Studdy Solution

STEP 1

What is this asking? We need to find how far the glider moves from the center and where it is a short time after letting go. Watch out! Remember, the glider starts at its *furthest* point, not the middle!

STEP 2

1. Find the amplitude2. Find the position at t=0.300 st = 0.300 \mathrm{~s}t=0.300 s

STEP 3

STEP 4

The formula that connects these is vmax=ω⋅Av_{max} = \omega \cdot Avmax​=ω⋅A, where ω\omegaω is the **angular frequency** and AAA is the **amplitude**.Think of ω\omegaω like how fast the glider is oscillating back and forth.

STEP 5

STEP 6

STEP 7

STEP 8

Now, we want to find the glider's position at t=0.300 st = 0.300 \mathrm{~s}t=0.300 s.Since the glider starts at its *maximum* position to the right (because it was pulled to the right), we can use the cosine function to describe its motion: x(t)=Acos⁡(ωt) x(t) = A \cos(\omega t) x(t)=Acos(ωt)

STEP 9

STEP 10

STEP 11

STEP 12

Part A: The amplitude of the oscillation is approximately 2292 cm2292 \mathrm{~cm}2292 cm.Part B: The glider's position at t=0.300 st = 0.300 \mathrm{~s}t=0.300 s is approximately 2292 cm2292 \mathrm{~cm}2292 cm to the right of the equilibrium position.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to find how far the glider moves from the center and where it is a short time after letting go. Watch out! Remember, the glider starts at its furthest point, not the middle!

1. Find the amplitude
2. Find the position at $t = 0.300 \mathrm{~s}$

The formula that connects these is $v_{max} = \omega \cdot A$ , where $\omega$ is the angular frequency and $A$ is the amplitude.
Think of $\omega$ like how fast the glider is oscillating back and forth.

Now, we want to find the glider's position at $t = 0.300 \mathrm{~s}$ .
Since the glider starts at its maximum position to the right (because it was pulled to the right), we can use the cosine function to describe its motion: $x(t) = A \cos(\omega t)$

Part A: The amplitude of the oscillation is approximately $2292 \mathrm{~cm}$ .
Part B: The glider's position at $t = 0.300 \mathrm{~s}$ is approximately $2292 \mathrm{~cm}$ to the right of the equilibrium position.