Respuesta :
Explanation:
It is given that,
Spring constant of the spring, k = 475 N/m
Mass of the block, m = 2.5 kg
Elongation in the spring from equilibrium, x = 5.5 cm
(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.
(b) Let [tex]\omega[/tex] is the angular frequency. It is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega=\sqrt{\dfrac{475}{2.5}}[/tex]
[tex]\omega=13.78\ rad/s[/tex]
(c) Let T is the period. It is given by :
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\pi}{13.78}[/tex]
T = 0.45 s
(d) Frequency,
[tex]f=\dfrac{1}{T}[/tex]
[tex]f=\dfrac{1}{0.45}[/tex]
f = 2.23 Hz
(e) Let v is the maximum value of the block's velocity. It is given by :
[tex]v_{max}=\omega\times A[/tex]
[tex]v_{max}=13.78\times 5.5\times 10^{-2}[/tex]
[tex]v_{max}=0.757\ m/s[/tex]
The value of acceleration is given by :
[tex]a=\omega^2A[/tex]
[tex]a=(13.78)^2\times 5.5\times 10^{-2}[/tex]
[tex]a=10.44\ m/s^2[/tex]
Hence, this is the required solution.
