Question3. A certain mass hangs from a spring above a table. It is released from a height of 0.9 metres above the table and falls to a height of 0.1 m above the table before reversing direction and bouncing back to 0.9 m . The mass continues to move in a periodic up and down motion. It takes 1.2 seconds for the mass to return to the same position each time. b) Write an equation which expresses the height $h$ as a function of $\sin \theta$ .

Studdy Solution

STEP 1

1. The motion of the mass is periodic and can be modeled using trigonometric functions.
2. The maximum height is 0.9 meters and the minimum height is 0.1 meters.
3. The period of the motion is 1.2 seconds.
4. The function to be derived is in terms of $\sin \theta$ .

STEP 2

1. Determine the amplitude and vertical shift of the motion.
2. Determine the angular frequency of the motion.
3. Write the equation for height $h$ as a function of $\sin \theta$ .

STEP 3

The amplitude $A$ of the motion is half the difference between the maximum and minimum heights.
$A = \frac{0.9 - 0.1}{2} = 0.4$
The vertical shift $D$ is the average of the maximum and minimum heights.
$D = \frac{0.9 + 0.1}{2} = 0.5$

STEP 4

The period $T$ is given as 1.2 seconds. The angular frequency $\omega$ is related to the period by the formula:
$\omega = \frac{2\pi}{T} = \frac{2\pi}{1.2} = \frac{5\pi}{3}$

STEP 5

The equation for height $h$ as a function of time $t$ using sine is:
$h(t) = A \sin(\omega t) + D$
Substitute the values of $A$ , $\omega$ , and $D$ :
$h(t) = 0.4 \sin\left(\frac{5\pi}{3} t\right) + 0.5$
Since we want $h$ as a function of $\sin \theta$ , we set $\theta = \frac{5\pi}{3} t$ , giving:
$h(\theta) = 0.4 \sin(\theta) + 0.5$
The equation expressing the height $h$ as a function of $\sin \theta$ is:
$h(\theta) = 0.4 \sin(\theta) + 0.5$

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