\(1)\\a)\\A=2\\T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2\pi}=1(s)\\f=\dfrac{1}{T}=1(Hz)\\\varphi=-\dfrac{\pi}{6}(rad)\\b)\\L=2A=4(cm)\\c)\\t=1(s)\Rightarrow x=2\cos\left(2\pi.1-\dfrac{\pi}{6}\right)=\sqrt{3}(cm)\\2)\\T=4(s)\Rightarrow \omega =\dfrac{2\pi}{T}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}(rad/s)\)

Ban đầu vật ở biên âm `=> varphi=pi(rad)`

\(\Rightarrow x=24\cos\left(\dfrac{\pi}{2}t+\pi\right)\\3)\\A=4(cm)\\\dfrac{T}{4}=1\\\Rightarrow T=4(s)\Rightarrow \omega =\dfrac{2\pi}{T}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}(rad/s)\)

Ban đầu vật ở biên âm `=> varphi=pi(rad)`

\(\Rightarrow x=4\cos\left(\dfrac{\pi}{2}t+\pi\right)\)

Bạn tham khảo:

Câu 1:

a) $\begin{cases} A = 2 (cm)\\\omega = 2\pi rad/s => T = \frac{2\pi}{\omega} = 1 (s) => f = \frac{1}{T} = 1 (Hz)\\\varphi = - \frac{\pi}{6} \end{cases}$

b) $\text{L = 2A = 2.2 = 4 (cm)}$

c) $\text{t = 1 (s) => x = 2cos(2$\pi$.1 - $\frac{\pi}{6}$) = $\sqrt{3}$}$ (cm) $\\$

Câu 2:

$\begin{cases} A = 24 (cm)\\T = 4 (s) => \omega = \frac{\pi}{2} rad/s\\\begin{cases} x_0 = -24 (cm)\\\text{chiều dương}\\ \end{cases} => \varphi = \pi\\ \end{cases}$ $\\$ => x = 24cos($\frac{\pi}{2}$.t + $\pi$) (cm)

Câu 3:

$\begin{cases} A = 4 (cm)\\T = 4 (s) => \omega = \frac{\pi}{2} rad/s\\\begin{cases} x_0 = -4 (cm)\\\text{chiều dương}\\ \end{cases} => \varphi = \pi \end{cases}$ $\\$ => x = 4cos($\frac{\pi}{2}$.t + $\pi$) (cm)

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