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 0160 \leq 1 \leq 6, where tt is in secands. Chech Desmos d(t)=10cos(2π3(t0.3))+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
To find the first time the weight reaches a specific distance (e.g., 50 cm), solve d(t)=50 d(t) = 50 .
10cos(2π3(t0.3))+50=50 10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) + 50 = 50
10cos(2π3(t0.3))=0 10 \cos \left(\frac{2 \pi}{3}(t-0.3)\right) = 0
cos(2π3(t0.3))=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(t0.3)=π2 \frac{2 \pi}{3}(t-0.3) = \frac{\pi}{2}
t0.3=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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