Download If a simple pendulum oscillates with an amplitude 50 mm and time

If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum
velocity is
(a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s
Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore
vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s [Option (b)].
The following question appeared in Kerala Engineering Entrance 2005 test paper:
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is
at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration.
Then its time period in seconds is
(a) 1/ 2π√3 (b) 2π√3 (c) 2π/√3 (d) √3/2π (e) √3/π
The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2
–y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) =
√3. Period T = 2π/ω = 2π/√3.
Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl
of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its
period of oscillation?
(a) 2π√(m/37rg) (b) 2π√(m/rg) (c) 12π√(r/g) (d) 2π√(r/g) (e) 2π√(37r/g)
The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the
usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a
circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r
since the length of the pendulum is measured from the centre of gravity of the bob. The point of
‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g)
given in (c).
What will be the period of oscillation of a simple pendulum of length 100 cm in a spaceship in a
geostationary orbit?
(a) 0 sec (b) infinite (c) 1 sec (d) 100 sec
Well, in any satellite orbiting the earth (in any orbit), the condition of weightlessness exists (effective
g = 0), the pendulum does not oscillate and the period therefore is infinite.
Consider the following question:
A simple pendulum is arranged using a small metallic bob of mass ‘m’and a light rubber cord of
length ‘L’ (on suspending the bob), area of cross section ‘A’ and Young’s modulus ‘Y’. [One should
use inextensible cord only for simple pendulum!]. When this unconventional pendulum is at rest
in its mean position, the bob is pulled slightly down and is released. Then, the period of the
vertical oscillation of the bob is (assuming that the size of the bob is negligible compared to the
length of the cord)
(a) 2π√2L/g (b) 2π√(mL/YA) (c) 2π√ (m/YAL) (d) 2π√ (L/g)
The period as usual is given by T = 2π√(m/k). Here ‘m’ is the same as the mass of the bob. The force
constant can be found by writing the expression for Young’s modulus (since it arises from the elastic
force in the cord): Y = FL/A(δL) where δL is the increase in the length of the cord on pulling the bob
down with a force F. Therefore, the force constant, F/(δL) = YA/L. On substituting this value, the
period is 2π√(mL/YA).
The following MCQ on simple harmonic motion may generate a little confusion in some of you:
A sphere of mass M is arranged on a smooth inclined plane of angle θ, in between two springs of
spring constants K1 and K2 . The springs are joined to rigid supports on the inclined plane and to
the sphere (Fig). When the sphere is displaced slightly, it executes simple harmonic motion. What
is the period of this motion?
(a) 2π[Mgsinθ/(K1-K2)]½ (b) 2π[M/{K1K2/(K1+K2)}]½ (c) 2π[Mgsinθ/(K1+K2)]½ (d) 2π[M/(K1+K2)]½ (e)
2π[(K1+K2)/M]½
You should note that gravity has no effect on the period of
oscillation of a spring-mass system since the restoring force is supplied by the elastic force in the
spring. (It can oscillate with the same period in gravity free regions also). So, whether you place the
system on an inclined plane or a horizontal plane, the period is the same and is determined by the
effective spring constant and the attached mass only. The effective spring constant is K1 + K2 since
both the springs try to enhance the opposition to the displacement of the mass. The period of
oscillation, as usual is given by, T = 2π√(Inertia factor/Spring factor) = 2π√[M/(K1 + K2)], given in
option (d).
The following two questions (MCQ) appeared in Kerala Engineering Entrance 2006 test paper:
(1)The instantaneous displacement of a simple harmonic oscillator is given by y = A cos(ωt + π/4).
Its speed will be maximum at the time
(a) 2π/ω (b) ω/2π (c) ω/π (d) π/4ω (e) π/ω
This question was omitted by a fairly bright student who got selected with a good rank. The question
setter used the term speed (and not velocity) to make things very specific and to avoid the possible
confusion regarding the sign. So what he meant is the maximum magnitude of velocity. The velocity
is the time derivative of displacement: v = dy/dt = -Aω(sin ωt + π/4). Its maximum magnitude equal
to Aω is obtained when ωt = π/4, from which t = π/4ω.
(2) A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time
period π/5 s. The maximum value of the force acting on the particle is
(a) 5 N (b) 4 N (c) 0.5 N (d) 0.3 N (e) 0.15 N
If you remember the basic expression for period in the form, T = 2π√(m/k) where ‘k’ is the force
constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since
‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement
(amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.
[If you remember that ω = √(k/m) you can arrive at the answer since T = 2π/ω].
Q1. A mass M hangs in equilibrium on a spring. M is made to oscillate about the equilibrium
position by pulling it down 10 cm and releasing it. The time for M to travel back to the
equilibrium position for the first time is 0.50 s. Which line, A to D, is correct for these
oscillations?
Sol
Amplitude /cm
Period /s
A
10
1.0
B
10
2.0
C
20
2.0
D
20
1.0
Amplitude /cm
Period /s
A
10
1.0
B
10
2.0
C
20
2.0
D
20
1.0
It is pulled down 10 cm therefore the maximum displacement from the equilibrium
(amplitude) is 10 cm. To travel from the displacement to the equilibrium position is a quarter
of the whole cycle (A to midpoint to-A to midpoint to A (again)). Therefore period is 2.0 s
Q2. A particle oscillates with undamped simple harmonic motion. Which one of the
following statements about the acceleration of the oscillating particle is true?
A
It is least when the speed is greatest.
B
It is always in the opposite direction to its velocity.
C
It is proportional to the frequency.
D
It decreases as the potential energy increases.
A
It is least when the speed is greatest. Speed is greatest when x = zero
(therough the centre of the swing). If x = zero acceleration = zero, so this is
true.
B
It is always in the opposite direction to its velocity. It is always in the
opposite direction to displacement.
C
It is proportional to the frequency. It is proportional to frequency squared.
D
It decreases as the potential energy increases. That would mean it decreased
as x went from zero to A. It is at a maximum when x = A and zero when x =
zero, so this is not true.
Q3. Which one of the following statements is true when an object performs simple harmonic
motion about a central point O?
A
The acceleration is always away from O.
B
The acceleration and velocity are always in opposite directions.
C
The acceleration and the displacement from O are always in the same
direction.
D
The graph of acceleration against displacement is a straight line.
-----
A
The acceleration is always away from O. Acceleration is sometimes away
from and somtimes towards O, but if it is towards then displacement is
away from - and vice versa!
B
The acceleration and velocity are always in opposite directions. Look at
the graph - this is not the case.
C
The acceleration and the displacement from O are always in the same
direction. No - they are in opposite directions.
D
The graph of acceleration against displacement is a straight line. True acceleration is directly proportional to the displacement (with a negative
constant of proportionality)
A particle of mass m executes simple harmonic motion in a straight line with amplitude A
and frequency f. Which one of the following expressions represents the total energy of the
particle?
A
2
2
mfA2
B
2
2
mf2A2
C
4
2
m2f2A
D
4
2
mf2A2
---------------------2
mfA2
A
2
B
2
C
4
2
m2f2A
D
4
2
mf2A2
2
mf2A2
Total energy is composed of kinetic energy and potential energy. When PE is zero all of the
energy is KE and vice versa. Therefore the maximum velocity can be used to work out the
energy
E = 1/2 mv2
= 1/2 m (2 fA)2
= 2m
2 2
f A2
Q5. A body moves with simple harmonic motion of amplitude A and frequency b/2 .
What is the magnitude of the acceleration when the body is at maximum displacement?
A
zero
B
4 2Ab2
C
Ab2
D
4 2Ab-2
--A
zero
B
4 2Ab2
C
Ab2
D
4 2Ab-2
At maximum displacement x = A.
acceleration = -(2 f)2 x
= -4
2 2
b A/4
2
= -b2A
as we are asked for the magnitude only the minus sign can be ignored.
Q6. A simple pendulum and a mass-spring system are taken to the Moon, where the
gravitational field strength is less than on Earth.
Which line, A to D, correctly describes the change, if any, in the period when compared with
its value on Earth?
period of pendulum
period of mass-spring system
A
decrease
decrease
B
increase
increase
C
no change
decrease
D
increase
no change
--
For the pendulum g is inversely proportional to T2. Therefore if g decreased the period would
increase.
The spring system's period does not depend on g. It depends on the mass (which is the same
wherever the body is) and the spring constant. The spring constant depends on the internal
arrangment of the spring's particles. That is unaltered by position in a graviational field.
Q7. A simple pendulum and a mass-spring system both have the same time period T at the
surface of the Earth. If taken to another planet where the acceleration due to gravity was half
that on Earth, which line, A-D, in the table gives correctly the new periods?
simple pendulum
mass-spring
A
T
B
T
C
D
-simple pendulum
mass-spring
A
T
B
T
C
D
For the pendulum g is inversely proportional to T2.
Therefore if g halved the period would increase (by a factor of root 2).
The spring system's period does not depend on g. It depends on the mass (which is the same
wherever the body is) and the spring constant. The spring constant depends on the internal
arrangment of the spring's particles. That is unaltered by position in a gravitational field.
Q8. Which one of the following statements is not true for a body vibrating in simple
harmonic motion when damping is present?
A
The damping force is always in the opposite direction to the velocity.
B
The damping force is always in the opposite direction to the
acceleration.
C
The presence of damping gradually reduces the maximum potential
energy of the system.
D
The presence of damping gradually reduces the maximum kinetic energy
of the system.
A
The damping force is always in the opposite direction to the velocity.
B
The damping force is always in the opposite direction to the
acceleration.
C
The presence of damping gradually reduces the maximum potential
energy of the system.
D
The presence of damping gradually reduces the maximum kinetic energy
of the system.
--
A damping force is a form of friction - it opposes movement and will therefore be in the
opposite direction to velocity - so A is true.
Acceleration and velocity are not always opposite (see the graphs below) , so B is NOT true.
A damping force will transfer kinetic energy into heat energy thus gradually reducing the
total energy of the SHM system - therefore C and D are true.
Q9. The frequency of a body moving with simple harmonic motion is doubled. If the
amplitude remains the same, which one of the following is also doubled?
A
the time period
B
the total energy
C
the maximum velocity
D
the maximum acceleration
-



Time period would halve as T = 1/f
Energy is related to v2 (KE = 1/2 mv2) - therefore enegy would in crease by a factor of root 2.
Max velocity would double as v and f are proportional.
Acceleration relates to square of frequency so that is incorrect. It would increase by a factor
of four.
Q10. The time period of a pendulum on Earth is 1.0 s. What would be the period of a
pendulum of the same length on a planet with half the density but twice the radius of Earth?
A
0.5 s
B
1.0 s
C
1.4 s
D
2.0 s
A
0.5 s
B
1.0 s
C
1.4 s
D
2.0 s
--
Mass Effect
Assume planet is a sphere - volume would be 8x that of earth (cubed relationship) if radius
was twice as great.
density = mass/volume therefore mass depends on the product of density and volume.
volume increases mass by a factor of eightbut density decreases it by a factor of 2. Therefore
the overall effect is that mass of the planet is four times that of Earth.
-4g
Radius effect
g has a inverse square relationship with radius, so doubling the radius will
decrease g by a factor of 4
Overall effect
Planet will havethe same g as Earth
T will be the same 1.0 s
Q11. Which one of the following statements always applies to a damping force acting on a
vibrating system?
A
It is in the same direction as the acceleration.
B
It is in the same direction as the displacement.
C
It is in the opposite direction to the velocity.
D
It is proportional to the displacement.
-A
It is in the same direction as the acceleration.
B
It is in the same direction as the displacement.
C
It is in the opposite direction to the velocity.
D
It is proportional to the displacement.
The damping force is a frictional force - it opposes motion - it is therefore ALWAYS in the
opposite direction to velocity.
Q12. Which one of the following statements concerning the acceleration of an object moving
with simple harmonic motion is correct?
A It is constant.
B
It is at a maximum when the object moves through the centre of the
oscillation.
C It is zero when the object moves through the centre of the oscillation.
D It is zero when the object is at the extremity of the oscillation.
-A It is constant.
B
It is at a maximum when the object moves through the centre of the
oscillation.
C It is zero when the object moves through the centre of the oscillation.
D It is zero when the object is at the extremity of the oscillation.
Look at the graph - acceleration varies sinusoidally so choice A is incorrect.
At the centre of the oscillation x=0 - when that is the case acceleration is also zero so choice
B is untrue and C is true.
At the 'extremity' displacement is a mximum and acceleration is also maximum - but in the
opposite direction.
Q13. When the length of a simple pendulum is decreased by 600 mm, the period of
oscillation is halved. What was the original length of the pendulum?
A
800 mm
B
1000 mm
C
1200 mm
D
1400 mm
-A
800 mm
B
1000 mm
C
1200 mm
D
1400 mm
T2 = constant x L
we now have to do a bit of manipulation and simple algebra...
T12/L1 = T22/L2
L1= T12L2 /T22
L1 = L
L2 = L - 600
T1 = T
T2 = 0.5 T
L = T2(L - 600)/(0.5 T)2
L = 4(L - 600)
3L = 2400
L = 800 mm
(and therefore length of the second was 200 mm)
Double check
root 800 = 28.28
root 200 = 14.14 (half the value)
Which one of the following gives the phase difference between the particle velocity and the
particle displacement in simple harmonic motion?
A
/4 rad
B
/2 rad
C
3 /4 rad
D
2
rad
-A
/4 rad
B
/2 rad
C
3 /4 rad
D
2
rad
Look at the graphs - the peaks of displacement and velocity are 'out' from each other by a
quarter of a cycle.
One cycle is 2
radians therefore they are 'out' by
/2 radians.
Q15. A body executes simple harmonic motion. Which one of the graphs, A to D, best shows
the relationship between the kinetic energy, Ek, of the body and its distance from the centre of
oscillation?
-KE relates to velocity squared.
When x = 0 velocity is a maximum - when x = A velocity = zero.
Therefore choice C is a non-starter.
KE = 1/2 mv2 = 2
2 2
f (A2 - x2)
= a constant value - a constant multiplied by x2
an x2graph curves upwards from the origin - you should know that from GCSE maths
The shape of the graph will therefore be that of choice B