Answer
given,
force constant
k = 6.5 N/m
Amplitude = A = 10 cm
a) using conserving of energy
[tex]E = \dfrac{1}{2}kA^2[/tex]
[tex]E = \dfrac{1}{2}\times 6.5 \times 0.1^2[/tex]
E = 32.5 m J
now,
[tex]E = \dfrac{1}{2}kA^2+ \dfrac{1}{2}mv^2[/tex]
[tex]E = \dfrac{E}{4}+ \dfrac{1}{2}mv^2[/tex]
[tex]m= \dfrac{3E}{4}\dfrac{2}{v^2}[/tex]
[tex]m= \dfrac{3\times 32.5 \times 10^{-3}}{4}\dfrac{2}{0.3^2}[/tex]
m =0.542 Kg
b) time period of the motion
[tex]T= 2\pi \sqrt{\dfrac{m}{K}}[/tex]
[tex]T= 2\pi \sqrt{\dfrac{0.542}{6.50}}[/tex]
T = 1.814 s
c) F = m a and also
F = k A
now, computing both the formula
[tex]a_{max} = \dfrac{k A}{m}[/tex]
[tex]a_{max} = \dfrac{6.58\times 0.1}{0.542}[/tex]
[tex]a_{max} =1.21\ m/s^2[/tex]
