7 A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the oscillation is started by elongating the spring 0.1m and giving the block a speed of 3 mis then the amplitude of the oscillation is:​

Respuesta :

(K+U)
i

=(K+U)
f



0+
2
1

kA
2
=
2
1

mv
2
+
2
1

kx
2



2
1

(6.50 N/m)(0.100 m)
2
=
2
1

m(0.300 m/s)
2
+
2
1

(6.50 N/m)(5.00×10
−2
m)
2


3.25×10
−2
J=
2
1

m(0.300 m/s)
2
+8.12×10
−3
J

giving m=
9.0×10
−2
m
2
/s
2

2(2.44×10
−2
J)

=
0.542 kg



(b) ω=
m
k



=
0.542 kg
6.50 N/m



=3.46 rad/s

Then, T=
ω


=
3.46 rad/s
2π rad

=
1.81 s



(c) a
max

=Aω
2
=(0.100 m)(3.46 rad/s)
2
=
1.20 m/s
2