Question

Just the answer please.
1: http://imgur.com/qVZ6TCg
2: http://imgur.com/AikiAaY
3: http://imgur.com/0AdRoQ4

Answer 1

Answer on Question #58930 – Math – Trigonometry

Question

For the simple harmonic motion equation $d = 5\sin \left(\frac{\pi}{4} t\right)$, what is the frequency?

If necessary, use the slash (/) to denote a fraction.

Solution

$$d = 5 \sin \left(\frac {\pi}{4} t\right) = a \sin (\omega t),$$ $$\omega = \frac {\pi}{4}.$$

Frequency is $f = \frac{\omega}{2\pi} = \frac{\pi}{4} \cdot \frac{1}{2\pi} = \frac{1}{8} = 1/8 \, s^{-1}$.

Answer: $1/8 \, s^{-1}$.

Question

Find a model for simple harmonic motion if the position at $t = 0$ is 0, the amplitude is 5 centimeters, and the period is 4 seconds.

$$d = 5 \sin (4 t)$$ $$d = 4 \sin (5 t)$$ $$d = 5 \cos \left(\frac {\pi}{2} t\right)$$ $$d = 5 \sin \left(\frac {\pi}{2} t\right)$$

Solution

The formula for simple harmonic motion is

$$d = A \sin (\omega t + \varphi),$$

where the amplitude is $A = 5$ centimeters, the period is $T = \frac{2\pi}{\omega} = 4$ seconds, hence $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} s^{-1}$.

It is given that $d(0) = 0 \Rightarrow A \sin (\omega \cdot 0 + \varphi) = 0 \Rightarrow A \sin (\varphi) = 0$, hence $\varphi = 0$ or $\varphi = \pi$.

Thus, $d = 5 \sin \left(\frac{\pi}{2} t\right)$.

Answer: $d = 5 \sin \left(\frac{\pi}{2} t\right)$.

Question

Find a model for simple harmonic motion if the position at $t = 0$ is 6, the amplitude is 6 centimeters, and the period is 4 seconds.

$$d = 4 \sin \left\{\frac {\pi}{3} t \right\}$$ $$d = 6 \cos (4 t)$$ $$d = 4 \sin (6 t)$$ $$d = 6 \cos \left\{\frac {\pi}{2} t \right\}$$

Solution

The formula for simple harmonic motion is

$$d = A \sin (\omega t + \varphi),$$

where the amplitude is $A = 6$ centimeters, the period is $T = \frac{2\pi}{\omega} = 4$ seconds, hence $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} s^{-1}$.

It is given that $d(0) = 6 \Rightarrow A \sin(\omega \cdot 0 + \varphi) = 6 \Rightarrow 6 \sin(\varphi) = 6$, hence $\varphi = \frac{\pi}{2}$.

Thus, $d = 6 \sin \left( \frac{\pi}{2} t + \frac{\pi}{2} \right) = 6 \cos \left( \frac{\pi}{2} t \right)$.

Answer: $d = 6 \cos \left( \frac{\pi}{2} t \right)$.

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