Respuesta :
Answer:
(a). [tex]x=2.00\sin(3.00\pi t)[/tex], hence proved
(b). The maximum speed is 18.8 cm/s.
(c). The maximum acceleration is 177.65 cm/s².
(d). The total distance is 12 cm.
Explanation:
Given that,
Amplitude = 2.00 cm
Frequency = 1.50 Hz
Given equation of position of the particle is
[tex]x=2.00\ sin(3.00\pit)[/tex]
(a). show that the position of the particle is given by
[tex]x=2.00\sin(3.00\pi t)[/tex]
We know the general equation of S.H.M
[tex]x=A\sin(\omega t[/tex]...(I)
At t =0, x = 0
On differentiating equation (I)
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=A\omega\cos(\omega t)[/tex]
At t = 0, the particle moving to the right
[tex]V=A\omega[/tex] > 0
Given statement is true.
The equation of position is
[tex]x=A\sin(\omega t)[/tex]
here, [tex]\Omega= 2\pi f[/tex]
Put the value in the equation
[tex]x=2.00\sin(2\times1.50\pi t)[/tex]
[tex]x=2.00\sin(3.00\pi t)[/tex]
Hence proved.
(b). We need to calculate the maximum speed
[tex]V=A\omega\cos(\omega t)[/tex]....(II)
At t = 0,
[tex]V_{max}=A\omega[/tex]
Put the value into the formula
[tex]V_{max}=2.00\times2\pi\times1.50[/tex]
[tex]V_{max}=6\pi[/tex]
[tex]V_{max}=18.8\ cm/s[/tex]
(c). We need to calculate the maximum acceleration
Using equation (II)
[tex]V=A\omega\cos(\omega t)[/tex]
On differentiating
[tex]a=\dfrac{dV}{dt}[/tex]
[tex]a=-A\omega^2\sin(\omega t)[/tex]
[tex]a_{max}\ when\ \sin(\omega t)\ is\ -1[/tex]
[tex]a_{max}=-A\omega^2\times-1[/tex]
[tex]a=A\omega^2[/tex]
[tex]a_{max}=2\times(3\pi)^2\approx 177.65 cm/s^2[/tex]
(d). We need to calculate the total distance traveled between t = 0 and t = 1.00 s
Using equation (II)
[tex]V=A\omega\cos(\omega t)[/tex]
On integration
[tex]\int{V}=\int_{t}^{t'}{A\omega\cos(\omega t)}[/tex]
Put the vale into the formula
[tex]\int{V}=\int_{0}^{1}{A\omega\cos(\omega t)}[/tex]
[tex]D=\int_{0}^{1}|6\pi\cos\left(3\pi t\right)|dt[/tex]
[tex]D=12\ cm[/tex]
Hence, (b). The maximum speed is 18.8 cm/s.
(c). The maximum acceleration is 177.65 cm/s².
(d). The total distance is 12 cm.
