Explanation:
Given that,
Spring constant of the spring, k = 6.5 N/m
Amplitude of simple harmonic motion, A = 11.5 cm
Speed of the block, v = 29 cm/s = 0.29 m/s
(a) When the block is halfway between its equilibrium position and the end point, x = 5.75 cm
The speed of oscillator at a given position is given by :
[tex]v=\sqrt{\dfrac{k}{m}(A^2-x^2)}[/tex]
[tex]m=\dfrac{k}{v^2}(A^2-x^2)[/tex]
[tex]m=\dfrac{6.5}{(0.29)^2}((11.5\times 10^{-2})^2-(5.75\times 10^{-2})^2)[/tex]
m = 0.766 kg
(b) The period of motion is given by :
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega=\sqrt{\dfrac{6.5}{0.766}}[/tex]
[tex]\omega=2.91[/tex]
[tex]T=\dfrac{2\pi}{2.91}[/tex]
T = 2.15 s
(c) The maximum acceleration of the block is given by :
[tex]a=\omega^2A[/tex]
[tex]a=(2.91)^2\times 11.5\times 10^{-2}[/tex]
[tex]a=0.973\ m/s^2[/tex]
Hence, this is the required solution.
