Answer:
Step-by-step explanation:
We have been given with the function
[tex]y=\frac{3}{2}cos{\frac{t}{2}}[/tex]
General form is [tex]y=acosbx[/tex]
Here, a is amplitude and b is the coefficient of x
Amplitude will be [tex]\frac{3}{2}[/tex]
Period of cos x is [tex]2\pi[/tex]
Whenever we need to find the period of any function we divide the general period of the function by coefficient of x
Hence, period of [tex]\frac{3}{2}cos{\frac{t}{2}}=\frac{2\pi}{\frac{1}{2}}=4\pi[/tex]
Range of cos x is [tex]-1\leq cost\leq1[/tex]
Divide each term by 2 we get [tex]\frac{-1}{2}\leq cos{\frac{t}{2}}\leq{\frac{1}{2}}[/tex]
Now, multiply each term by 3/2 we get:
[tex]\frac{-3}{4}\leq \frac{3}{2}cos{\frac{t}{2}}\leq \frac{3}{4}[/tex]
The period of the oscillation is 12.57 s.
The amplitude of cosine function is 3/2 m.
The range of the cosine function is -³/₄ ≤ ³/₂ Cos t/₂ ≤ ³/₄.
The general wave equation is given by the following formula;
y = -Acosωt
where;
From the given wave equation, ( y = 3/2 cos t/2), the amplitude is 3/2 m.
The period of the oscillation is calclated as follows;
cos ωt = cos t/2
cos 2πft = cos t/2
2πf = ¹/₂
f = 1/4π
T = 1/f = 4π
T = 4(3.142) = 12.57 s
Range of Cos x, - 1 ≤ x ≤ 1
y -A cos t/2 = ¹/₂ (-ACost)
y = ¹/₂(-³/₂) ≤ ³/₂ Cos t/₂ ≤ (¹/₂ x ³/₂)
y = -³/₄ ≤ ³/₂ Cos t/₂ ≤ ³/₄
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