Question 4
A spring is hung
vertically from a fixed point. A mass M is hung from the other end of the
spring, as illustrated in Fig. 3.1.
Fig. 3.1
The mass is displaced
downwards and then released. The subsequent motion of the mass is simple
harmonic.
The variation with
time t of the length L
of the spring is shown in Fig. 3.2.
Fig. 3.2
(a)
State:
(i)
one time at which the mass is moving with maximum speed [1]
(ii)
one time at which the spring has maximum elastic potential energy.
[1]
(b)
Use data from Fig. 3.2 to determine, for the motion of the mass:
(i)
the angular frequency ω [2]
(ii)
the maximum speed [2]
(iii)
the magnitude of the maximum acceleration. [2]
(c)
The mass M is now suspended from two springs, each identical to that
in Fig. 3.1, as shown in Fig. 3.3.
Fig. 3.3
Suggest and explain
the change, if any, in the period of oscillation of the mass. A numerical
answer is not required. [2]
[Total: 10]
Reference: Past Exam Paper – June 2019 Paper 42 Q3
Solution:
(a)
(i)
{In s.h.m., the mass moves with maximum speed
at the equilibrium position – here, this corresponds to any time where L = 12
cm}
0.10 s or 0.30 s or 0.50 s or 0.70 s or 0.90 s
(ii)
{Maximum elastic potential energy occurs when
the spring is at its longest – this corresponds to the lowest position.}
0 or 0.40 s or 0.80 s
{At the shortest length, the spring is
compressed and in this situation, the mass also has GPE.}
(b)
(i)
ω = 2π / T
ω = 2π / 0.40
ω = 16 rad s-1
(ii)
{The
maximum displacement x0 (amplitude) is 2.5 cm.}
v0 = ωx0
v0 =
15.7 × 2.5 × 10-2
v0 =
0.39 m s-1
(iii)
{Acceleration
a = ω2x
The
acceleration a is greatest when displacement x is maximum (that is, amplitude)}
a0 = ω2x0
a0 =
(15.72 × 2.5 × 10-2)
a0 =
6.2 m s-2
or
a0 = ωv0
a0 =
15.7 × 0.39
a0 =
6.2 m s-2
(c)
The period of oscillation decreases as the acceleration is
greater (for any given extension).
{A parallel
combination of springs has a greater spring constant k.
From Hooke’s
law: F = ke
With a
greater spring constant, the restoring force F is greater (for any given
extension).
With the 2
springs, when the mass M is positioned (without oscillating), the springs are
extended. But this extension is smaller than before as the spring constant is
now greater. This position is the equilibrium position. So, using 2 springs in
parallel results in a new equilibrium position of the mass.
The
amplitude of the oscillations is unconnected to the number of springs used.
To compare
with the case of a single spring, for any given amplitude (similar in both
case), the mass would have greater energy with 2 springs (elastic energy = ½ kx2
and k is now greater while x is taken to be the same in both case). As the mass
moves towards the equilibrium position, this energy is converted into KE. So, a
greater amount of elastic potential energy means that more energy is converted
into KE, and so, the mass would have a greater maximum speed.}
