Explanation:
It is given that,
Mechanical energy of the block- spring system, E = 3.27 J
Amplitude of the system, A = 10.3 cm = 0.103 m
Maximum speed, v = 1.1 m/s
(a) Mechanical energy of the block spring system is given by :
[tex]E=\dfrac{1}{2}kA^2[/tex]
[tex]k=\dfrac{2E}{A^2}[/tex]
[tex]k=\dfrac{2\times 3.27}{(0.103)^2}[/tex]
k = 616.45 N/m
(b) Velocity is maximum at the equilibrium position. The mechanical energy at the equilibrium position is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]m=\dfrac{2E}{v^2}[/tex]
[tex]m=\dfrac{2\times 3.27}{(1.1)^2}[/tex]
m = 5.4 kg
(c) The frequency of oscillation is :
[tex]f=2\pi\sqrt{\dfrac{m}{k}}[/tex]
[tex]f=2\pi\times \sqrt{\dfrac{5.4}{616.45}}[/tex]
f = 0.58 Hz
Hence, this is the required solution.
