Answer :
Answer:
a) [tex]A=0.165m[/tex]
b) 29
c) [tex]K.E_{max}=3.0\ \textup J[/tex]
d) [tex]V=1.713m/s[/tex]
Explanation:
mass, m =0.96 g
k = 220 N/m
Total energy, E = 3.0 J
Now,
a) [tex]E=\frac{1}{2}kA^2[/tex]
where, A is the amplitude
on substituting the values, we get
[tex]3.0=\frac{1}{2}\times 220\times A^2[/tex]
or
[tex]A^2=\frac{3\times2}{220}[/tex]
or
[tex]A^2=0.02727[/tex]
or
[tex]A=0.165m[/tex]
b) Time period (T) is given as:
[tex]T=2\pi\sqrt\frac{m}{k}[/tex]
on substituting the values,we get
[tex]T=2\pi\sqrt\frac{0.96}{220}[/tex]
or
[tex]T=0.415s[/tex]
thus, number of oscillations (N) in 12s will be
[tex]N=\frac{12}{0.415}=28.91\approx29\ \textup{oscillations}[/tex]
c)Maximum K.E = total mechanical energy
thus,
[tex]K.E_{max}=3.0\ \textup J[/tex]
d)The angular frequency (ω) is given as:
[tex]\omega=\sqrt\frac{k}{m}[/tex]
on substituting the values,we get
[tex]\omega=\sqrt\frac{220}{0.96}[/tex]
or
[tex]\omega=15.13 s^{-1}[/tex]
Now, the speed (V) in SHM is calculated as;
[tex]V=\omega\sqrt{A^2-x^2}[/tex]
for x = 0.12m
we get
[tex]V=15.13\times \sqrt{0.165^2-0.12^2}[/tex]
or
[tex]V=1.713m/s[/tex]