Oscillations Test 7

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Oscillations Test 7

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Question 1
1 / -0

The pendulum of a wall clock executes

A
a translatory motion

B
a rotary motion

C
a random motion

D
an oscillatory motion

Solution

The bob of a pendulum moves in such a way that it repeats its positions several time but alternately the time gap between these positions is always equal which proves that the motion of a pendulum is oscillatory.
Question 2
1 / -0

In a simple pendulum the restoring force is due to

A
The tangential component of the gravitational force

B
The radial component of the gravitational force

C
The tangential component of the tension in string

D
The radial component of the tension in string

Solution

In a simple pendulum the restoring force is due to the tangential component of the gravitational force because on applying torque equation on the radial forces the effect is nullified due to it passing from centre & only tangential component remains.
Question 3
1 / -0

The kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. Then, the displacement of the body is x percent of the amplitude, where x is

A
33

B
50

C
67

D
87

Solution

We know that PE + KE = TE = constant
Hence at the extremis TE = PE = ½ kz²
Where z is amplitude and k is shm constant.
Thus when KE = ⅓ PE
We get PE = ¾ TE = ½ kz²
Hence we get ½ kx² = ¾ ½ kz²
We get x/z = √3/4
= 1.73 / 2
= .87
Thus we get x is 87 percent of the amplitude.
Question 4
1 / -0

If the length of a simple pendulum is increases by 1percent, then the new time-period

A
Decreases by 0.5 percent

B
Increases by 0.5 percent

C
Increases by 1 percent

D
Increases by 2.0 percent

Solution

Time period, T=2π√l/g
=>T∝√l
=>ΔT/T=1/2 Δl/l
Δl/l=1%
Therefore, ΔT/T=1/2x1%=0.5%
Question 5
1 / -0

A body is executing S.H.M When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocities of the body are 10 cm per sec and 8 cm per sec. Then the time period of the body is

A
2 π/2 sec

B
π sec

C
3 π/2 sec

D
2 π sec

Solution

v=ω√A²−x2
⇒8=ω √A²−25
&10=ω√A²−16
⇒(8/10)2= A²−25/ A²−16
⇒16/25= A²−25/ A²−16
⇒16A²−256=25A²−625
⇒9A²=369
A²=41
⇒8=ω41−25
⇒8=4ω
∴ω=2=T2π
T=π sec
Question 6
1 / -0

Oscillatory motion is

A
combination of translation and rotation

B
periodic to and fro motion

C
rotation about a point

D
none

Solution

Oscillatory motion is periodic to and fro motion. By definition
Question 7
1 / -0

In a damped system

A
energy of the system never decreases

B
energy of the system is continuously dissipated

C
energy of the system is continuously refreshed

D
energy of the system continuously increases
Question 8
1 / -0

A spring with spring constant k when stretched through 1 cm has a potential energy U. If it is stretched by 4 cm, the potential energy will become

A
4U

B
16U

C
8U

D
32U

Solution
Question 9
1 / -0

Two pendulums have time periods T and 5T/4. They are in phase at their mean positions at some instant of time. What will be their phase difference when the bigger pendulum completes one oscillation?

A
30⁰

B
45⁰

C
60⁰

D
90⁰

Solution
Question 10
1 / -0

The ratio of maximum acceleration to maximum velocity of a particle performing S.H.M is equal to

A
Square of amplitude

B
Square of angular velocity

C
Amplitude

D
Angular velocity

Solution

Maximum acceleration = w2A
Maximum velocity = wA
Ratio = w2A/wA = w
w = angular velocity
A = amplitude
Question 11
1 / -0

Periodic motion is

A
motion that involves coming to a mean position at random intervals of time

B
motion that repeats itself at regular intervals of time

C
motion that does not repeat itself at regular intervals of time

D
motion that repeats itself at irregular intervals of time
Question 12
1 / -0

Damping is due to

A
electrostatic forces

B
resistive forces like air drag, friction etc.

C
reaction forces

D
conservative forces like gravity

Solution

Damping is due to resistive forces like air drag, friction etc.
Question 13
1 / -0

Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulum will be one behind the other is

A
2T/5

B
T/3

C
T/4

D
4T/3

Solution

T1=2π√1/g=T, T₂=2π√16/g=4T
(ω1−ω2)t=2π
(2π/T−2π/4T)t=2π
t=4T/3
Question 14
1 / -0

The period of a simple pendulum is doubled when

A
The mass of the bob and the length of the pendulum are doubled

B
The mass of the bob is doubled

C
Its length is increased four times

D
Its length is doubled
Question 15
1 / -0

A particle is executing S.H.M with an amplitude of 4 cm and time period 12 sec. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is T₁..The time taken from this displaced position of 2 cm to reach the extreme position is T2. Therefore, T₁/ T₂ will be

A
1/ 3

B
1/2

C
41276.0

D
1
Question 16
1 / -0

Period of a periodic motion is the ________ interval of time after which the motion is __________.

A
smallest, repeated

B
mean, not repeated

C
largest, repeated

D
average, repeated

Solution

Explanation:Time period of an oscillatory motion is the smallest interval of time for one complete oscillation.As when we apply the force on the bob,bob moves from mean position to extreme position then form extreme position to mean position and then again from mean to extreme in opposite direction and at last come to mean position.The time taken by bob in this whole process is called period of oscillation.
Question 17
1 / -0

In simple models damping force is

A
directly proportional to velocity

B
inversely proportional to acceleration

C
inversely proportional to velocity

D
directly proportional to acceleration
Question 18
1 / -0

For all practical purposes, the motion of a simple pendulum is SHM,

A
Only if the maximum angle which the string makes with the vertical is less than I

B
Only if the length of its string is at least one meter

C
Only if the maximum angle which its string makes with the vertical is less than 342.

D
None of these

Solution

For all practical purposes, the motion of a simple pendulum is SHM only if the maximum angle which the string makes with the vertical is less than 1 because for angles less than 1° sinθ = θ.
Question 19
1 / -0

Whose motion is an ideal case of SHM?

A
An electric dipole placed stably in a uniform electric field and then given a very small regular displacement

B
A simple pendulum with very small angle of swing

C
Shadow of a point mass having uniform circular motion, on the vertical diameter

D
none

Solution

Simple pendulum is not a case of ideal SHM as it is mention in NCERT.
Shadow of a point mass having uniform circular motion on the horizontal diameter is a case of SHM but not on the vertical diameter.
Question 20
1 / -0

If the period of oscillation of mass M suspended from a spring is one second, then the period of mass 4 M will be

A
1/ 4s

B
4s

C
1/ 2s

D
2s

Solution

T=2*3.14*√m/√k=1 s (given)
T’=2*3.14*√4m/√k=2*3.14*2√m/√k=2*1s
Therefore T’=2s
Question 21
1 / -0

Frequency of the periodic motion is

A
the reciprocal of the period

B
twice the reciprocal of the period

C
half the reciprocal of the period

D
thrice the reciprocal of the period
Question 22
1 / -0

Damped natural frequency is

A
same as natural frequency

B
lower than natural frequency

C
higher than natural frequency

D
none of the above

Solution

If a resonant mechanical structure is set in motion and left to its own devices, it will continue to oscillate at a particular frequency known as its natural frequency, or "damped natural frequency". This will be a little lower in frequency than the resonant frequency, which is the frequency it would assume if there were no damping. The resonant frequency is also called the "undamped natural frequency”
Question 23
1 / -0

A rubber ball with water, having a small hole in its bottom is used as the bob of a simple pendulum. The time-period of such a pendulum:

A
Increases with time

B
First increases and then decreases finally having same value as at the beginning

C
Decreases with time

D
Is a constant

Solution

When rubber ball completely filled with water its centre of gravity will be at its centre, as water will fall through hole its COG will shift towards lower side leading to increase in length of pendulum and thus T, when very small amount of water will be left in rubber ball its COG will again shifts upward causing decrease in length and thus T, and finally when rubber ball becomes empty its COG will be at its centre and T will remains same as earlier.
Question 24
1 / -0

A particle executes S.H.M. having time period T, then the time period with which the potential energy changes is:

A
2T

B
T

C
Infinite

D
T/2
Question 25
1 / -0

The length of the second’s pendulum on the surface of earth is 1m. The length of second’s pendulum on the surface of moon, where g is 1/6th the value on earth is

A
6m

B
1/ 36m

C
36m

D
1/ 6m

Solution

Time period remains same since
T = 2π √(L/g)
So, from proportionality we can solve
Question 26
1 / -0

Any periodic function can be expressed as a.

A
superposition of sine and cosine functions of different time periods with suitable coefficients

B
superposition of exponential functions of real time periods with suitable coefficients

C
superposition of polynomial functions of different time periods with suitable coefficients

D
superposition of sine and cosine functions of same time period with suitable coefficients

Solution

All periodic functions can, in principle, be represented as a sum of sine and cosine functions, since they form what's known as a complete set.
Question 27
1 / -0

In forced oscillations apart from acceleration forces, damping and spring forces there is

A
damping force

B
an external exciting force that changes the energy

C
restoring force

D
gravity force
Question 28
1 / -0

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, the time period becomes 5T/3. then the ratio of m/M is

A
16/9

B
25/ 16

C
3/9

D
25/22

Solution
Question 29
1 / -0

The length of a second’s pendulum decreases by 0.1percent, then the clock

A
Loses 43.2 seconds per day

B
Loses 13.5 seconds per day

C
Loses 7 seconds per day

D
Gains 43.2 seconds per day

Solution

L’=1.001L
T=2π√L/9=1s
T’=2π√L’/g=2π√1.001/g=Tx√1.001=1.0005xT
T’=1.0005s
Error in time per day=1.0005x60x60x24-1x60x60x24
Error=86443.2-86400
Error=43.2s
Question 30
1 / -0

If the maximum velocity and acceleration of a particle executing S.H.M are equal in magnitude, the time period will be

A
6.28 s

B
1.57 s

C
3.14 s

D
12.56 s

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Oscillations Test 7

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Question 1

a translatory motion

a rotary motion

a random motion

an oscillatory motion

Solution

Question 2

The tangential component of the gravitational force

The radial component of the gravitational force

The tangential component of the tension in string

The radial component of the tension in string

Solution

Question 3

33

50

67

87

Solution

Question 4

Decreases by 0.5 percent

Increases by 0.5 percent

Increases by 1 percent

Increases by 2.0 percent

Solution

Question 5

2 π/2 sec

π sec

3 π/2 sec

2 π sec

Solution

Question 6

combination of translation and rotation

periodic to and fro motion

rotation about a point

none

Solution

Question 7

energy of the system never decreases

energy of the system is continuously dissipated

energy of the system is continuously refreshed

energy of the system continuously increases

Question 8

4U

16U

8U

32U

Solution

Question 9

300

450

600

900

Solution

Question 10

Square of amplitude

Square of angular velocity

Amplitude

Angular velocity

Solution

Question 11

motion that involves coming to a mean position at random intervals of time

motion that repeats itself at regular intervals of time

motion that does not repeat itself at regular intervals of time

motion that repeats itself at irregular intervals of time

Question 12

electrostatic forces

resistive forces like air drag, friction etc.

reaction forces

conservative forces like gravity

Solution

Question 13

2T/5

30⁰

45⁰

60⁰

90⁰