Question

A pendulum is swinging back and forth. After ttt seconds, the horizontal distance from the bob to the place where it was released is given by \qquad H(t) = 7 - 7 \cos \left(\dfrac{2\pi (t-2)}{20}\right)H(t)=7−7cos( 20 2π(t−2) ​ )H, left parenthesis, t, right parenthesis, equals, 7, minus, 7, cosine, left parenthesis, start fraction, 2, pi, left parenthesis, t, minus, 2, right parenthesis, divided by, 20, end fraction, right parenthesis. How often does the bob cross its midline? Give an exact answer

Answer 1

The bob of a swinging pendulum crosses its midline every 10 seconds.

Step-by-step explanation:

The equation H(t) = 7 - 7 cos((2π(t-2))/20) represents the horizontal distance from the bob of a swinging pendulum to the point where it was released. To find how often the bob crosses its midline, we need to determine when H(t) = 0. This happens when cos((2π(t-2))/20) = 1, which occurs when the argument of the cosine function is a multiple of 2π. So, we set (2π(t-2))/20 = 2πn, where n is an integer. Solving for t gives us t = 10n + 2. Therefore, the bob crosses its midline every 10 seconds, starting from t = 2.

Answer 2

Every 10 seconds

Step-by-step explanation:

The bob crosses its midline whenever cos(2π(t-2)/20)=0.

Since cosθ=0 when θ=±π/2 + 2πn, we can find then the bob crosses its midline by solving:

2π(t-2)/20 = ±π/2 + 2πn

t-2 = ±5+20n

The solutions are when t = -3, 7, 17, 27, 37, ...

Therefore, the bob passes its midline every 10 seconds.