10. A load of mass 0.5 kg hangs from a spring of force constant 10 Nm-1. The
mass is pulled down 0.05 m from its equilibrium position and then released.
(a) What is the distance between the two most widely separated positions
of the mass ? (b) How long does it take to traverse this distance ?
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Answer:Spring constant, k=1200m
−1
Mass, m=3kg
Displacement, A=2.0cm=0.02m
(i) Frequency of oscillation v, is given by the relation:
v=
T
1
=
2π
1
m
k
where, T is time period
∴v=
2×3.14
1
3
1200
=3.18Hz
Hence, the frequency of oscillations is 3.18 cycles per second.
(ii) Maximum acceleration (a) is given by the relation:
a=ω
2
A
where,
ω= Angular frequency =
m
k
A = maximum displacement
∴a=
m
k
A=
3
1200×0.02
=8ms
−2
Hence, the maximum acceleration of the mass is 8.0m/s
2
(iii) Maximum velocity, v
max
=Aω
=A
m
k
=0.02×
3
1200
=0.4m/s
Hence, the maximum velocity of the mass is 0.4 m/s.
Explanation:
hope it helps
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