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Math  /  Trigonometry

Question\#3 A weight atrached to the end of a long spring hanging above the ground is bouncing up and down As it bounces, its distance from the floor varies sinuscidally with time (assume no friction is present in the spring) A stopwatch is used to measure is height above the foor as a function of time. When the stopvatch reads 0.3 s , the weight first reaches a ligh point C0 cmC 0 \mathrm{~cm} above the foor. The next low point, at 40 cm above the floor, occurs at 1.8 s . a. Oraw a shetch to dlustrate dd, the spring's distance from the foor in centimetres, over the interval 0≤1≤60 \leq 1 \leq 6, where tt is in secands. Chech Desmos d(t)=10cos⁡(2π3(t−0.3))+50\begin{array}{l} d(t)=10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right)+50 \end{array} c. What is the distance the fioor for the first time?

Studdy Solution

STEP 1

1. The weight's motion is sinusoidal, described by a cosine function.
2. The high point is at 60 cm above the floor at t=0.3 t = 0.3 seconds.
3. The low point is at 40 cm above the floor at t=1.8 t = 1.8 seconds.
4. The function d(t)=10cos⁡(2π3(t−0.3))+50 d(t) = 10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) + 50 describes the distance from the floor.

STEP 2

1. Understand the sinusoidal function and its parameters.
2. Sketch the function over the given interval.
3. Determine the first time the weight reaches a specific distance from the floor.

STEP 3

The function d(t)=10cos⁡(2π3(t−0.3))+50 d(t) = 10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) + 50 represents a cosine wave with: - Amplitude: 10 cm - Period: 2π2π3=3 \frac{2\pi}{\frac{2\pi}{3}} = 3 seconds - Phase shift: 0.3 seconds to the right - Vertical shift: 50 cm

STEP 4

To sketch the function: - Start at the high point (60 cm) at t=0.3 t = 0.3 seconds. - The next low point (40 cm) occurs at t=1.8 t = 1.8 seconds. - The function will complete one full cycle every 3 seconds.

STEP 5

To find the first time the weight reaches a specific distance (e.g., 50 cm), solve d(t)=50 d(t) = 50 .
10cos⁡(2π3(t−0.3))+50=50 10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) + 50 = 50
10cos⁡(2π3(t−0.3))=0 10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) = 0
cos⁡(2π3(t−0.3))=0 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) = 0
The cosine function equals zero at π2+kπ \frac{\pi}{2} + k\pi , where k k is an integer.
2π3(t−0.3)=π2 \frac{2 \pi}{3}(t-0.3) = \frac{\pi}{2}
t−0.3=34 t-0.3 = \frac{3}{4}
t=34+0.3=1.05 t = \frac{3}{4} + 0.3 = 1.05
The first time the weight is 50 cm above the floor is:
1.05 seconds \boxed{1.05 \text{ seconds}}

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