Answer:
(a) 45 cm (b) 3.305 rad/sec (c) 1.4875 m/sec
Explanation:
It is given that the motion is simple harmonic and it takes 1.90 sec to complete a cycle
The frequency is defined as the reciprocal of time taken to complete one cycle so [tex]f=\frac{1}{1.90}=0.5263\ Hz[/tex]
(a) As it is given that the height of each bounce above the equilibrium positon is 45 cm so amplitude will be 45 cm
(b) The angular frequency ω is given by [tex]\omega =2\pi f=2\times 3.14\times 0.5263=3.3051[/tex] rad/sec
(c) The maximum speed [tex]V_{max}[/tex] is given by [tex]V_{max}=\omega A[/tex] where ω is angular frequency of the motion and A is the amplitude So [tex]V_{max}=3.305\times 45\times 10^{-2}=1.4875\ m/sec[/tex]
