Given:
The mass, m=0.5 kg
The velocity is given by,
[tex]v_x(t)=-(3.6\text{ cm/s})\sin \lbrack(4.71\text{ rad/s})t-\frac{\pi}{2}\rbrack\text{ }\rightarrow\text{ (i)}[/tex]To find:
A) Period
B) Amplitude
C) Maximum acceleration of the mass.
D) Force constant of the spring.
Explanation:
The displacement of a mass in simple harmonic motion is given by the equation,
[tex]x=A\cos \lbrack\omega t-\phi\rbrack[/tex]Where A is the amplitude, ω is the angular velocity, t is the time, and φ is the phase difference.
By differentiating the above equation, we get the velocity of the object in simple harmonic motion.
Thus,
[tex]v_x(t)=-A\omega\sin \lbrack\omega t-\phi\rbrack\text{ }\rightarrow\text{ (ii)}[/tex]On comparing equation (i) and equation (ii),
[tex]\begin{gathered} A\omega=3.6\text{ cm/s} \\ \omega=4.7\text{ rad/s} \\ \phi=\frac{\pi}{2} \end{gathered}[/tex]A)
The period is given by,
[tex]T=\frac{2\pi}{\omega}[/tex]On substituting the known values,
[tex]\begin{gathered} T=\frac{2\pi}{4.7} \\ =1.34\text{ s} \end{gathered}[/tex]B)
We know,
[tex]A\omega=3.6\text{ cm/s}[/tex]On substituting the known values,
[tex]\begin{gathered} A\times4.7=3.6 \\ \Rightarrow A=\frac{3.6}{4.7} \\ =0.77\text{ cm} \end{gathered}[/tex]C)
The maximum acceleration of a mass in SHM is given by,
[tex]a_{\max }=A\omega^2[/tex]On substituting the known values,
[tex]\begin{gathered} a_{\max }=0.77\times10^{-2}\times4.7^2 \\ =0.17\text{ m/s}^2 \end{gathered}[/tex]D)
The period of the spring is given by,
[tex]T=2\pi\sqrt[]{\frac{m}{k}}[/tex]Where k is the force constant of the spring.
On rearranging the above equation,
[tex]\begin{gathered} \frac{T^2}{4\pi^2}=\frac{m}{k} \\ \Rightarrow k=\frac{m\times4\pi^2}{T^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} k=\frac{0.5\times4\pi^2}{1.34^2} \\ =11\text{ N/m} \end{gathered}[/tex]Final answer:
A) The period is 1.34 s
B) The amplitude is 0.77 cm
C) The maximum acceleration of the mass is 0.17 m/s²
D) The force constant is 11 N/m
