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Angular simple harmonic motion - class-XI

Description: angular simple harmonic motion
Number of Questions: 49
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Tags: physics oscillatory motion oscillations
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A large box is accelerated up the inclined plane with an acceleration a and pendulum is kept vertical (Somehow by an external agent) as shown in figure.Now if the pendulum is set free to oscillate from such position, then what is the tension in the string immediately after the pendulum is set free? (mass of $500m$)

  1. $mg$

  2. $ma _{o} \sin\theta$

  3. $\left( m g + m a _ { 0 } \sin \theta \right)$

  4. Zero


Correct Option: C

 The time period of oscillation of a torsional pendulum of moment of inertia I is

  1. $T =2 \pi \sqrt{I/k}$

  2. $T =2 \pi \sqrt{I/2k}$

  3. $T =2 \pi \sqrt{2I/k}$

  4. $T =2 \pi \sqrt{I/4k}$


Correct Option: A
Explanation:

The time period of oscillations of a torsional pendulum is $T =2 \pi \sqrt(I/k)$

The correct option is (a)

A bullet of mass $'m'$ hits a pendulum bob of mass $'2m'$ with a velocity $'v'$ and comes out of the bob with velocity $v/2$. Length of the pendulum is $2$ meter and $g=10 ms^{-2}$. The minimum value of $'v'$ for the bullet so that the bob may complete one revolution in the verticle is

  1. $40 ms^{-1}$

  2. $2.20 ms^{-1}$

  3. $3.15 ms^{-1}$

  4. $10 ms^{-1}$


Correct Option: A

Time period of a disc about a tangent parallel to the diameter is same as the time period of a simple pendulum. The ratio of radius of disc to the length of pendulum is :

  1. $\dfrac { 1 } { 4 }$

  2. $\dfrac { 4 } { 5 }$

  3. $\dfrac { 2 } { 3 }$

  4. $\dfrac { 1 } { 2 }$


Correct Option: C

A pendulum of mass $m$ hangs from a support fixed to a trolley. The direction of the string (i.e.., angle $\theta$) when the trolley rolls up a plane of inclination $\alpha$ with acceleration $'a'$ is

  1. Zero

  2. $\tan^{-1} \alpha$

  3. $\tan^{-1}\dfrac{a+g \sin \alpha}{g \cos \alpha}$

  4. $\tan^{-1}\dfrac{a}{g}$


Correct Option: C

The oscillations of a pendulum slow down due to

  1. the force exerted by air and friction at the support

  2. the foce exerted by air only

  3. the forces exerted by friction at the support only

  4. none of these


Correct Option: A
Explanation:

The pendulum slows down due to friction force exerted by air and friction at the support.

The pendulum of a certain clock has time period $2.04 s$. How fast or slow does the clock run during $24$ hour?

  1. $28.8$ minutes slow

  2. $28.8$ minutes fast

  3. $14.4$ minutes fast

  4. $14.4$ minutes slow


Correct Option: A

A string of simple pendulum can bear maximum tension that is twice the weight of the bob. What is the maximum angle $(\theta)$ with which it can oscillate?

  1. $0^0$

  2. $45^0$

  3. $60^0$

  4. $90^0$


Correct Option: B

A bob is suspended from an ideal string of length $l$. Now it is pulled to a side through $60^{o}$ to vertical and rotates along a horizontal circle. Then its period of revolution is

  1. $ 2\pi \sqrt{ l/g }$

  2. $ \pi \sqrt{ l/2g }$

  3. $\pi\sqrt{ 2l/g }$

  4. $\pi\sqrt{ l/g }$


Correct Option: C

Two simple pendulums have time period $4\ s$ and $5\ s$ respectively. If they started simultaneously from the mean positive in the same direction, then the phase difference between them by the time the larger one completes one osicillation is

  1. $\dfrac {\pi}{6}$

  2. $\dfrac {\pi}{3}$

  3. $\dfrac {\pi}{2}$

  4. $\dfrac {\pi}{4}$


Correct Option: A

Write the torque equation for the bob of a pendulum if it makes an angle of $\theta$ with the vertical and I is the moment of inertia of the bob w.r.t the point of suspension

  1. $I \dfrac{d^2 \theta}{dt^2}=mgL \cos \theta$

  2. $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

  3. $I \dfrac{d^2 \theta}{dt^2}=mgL \tan \theta$

  4. $I \dfrac{d^2 \theta}{dt^2}=mg \sin \theta$


Correct Option: B
Explanation:

Taking the torque about the point of suspension, we can write $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

The correct option is (b)

One end of spring of spring constant k is attached to the centre of a disc of mass m and radius R and the other end of the spring connected to a rigid wall. A string is wrapped on the disc and the end A of the string is pulled through a distance a and then released.
The disc is placed on a horizontal rough surface and there is no slipping at any contact point What is the amplitude of the oscillation of the centre of the disc?

  1. a

  2. 2a

  3. a/2

  4. none of these.


Correct Option: C
Explanation:

Displacement of the topmost point of the disc = a.
Disc undergoes rolling without slipping.
Hence the displacement of the centre of the disc = a/2
Thus the amplitude of the oscillation of the centre of the disc = a/2
Hence (C) is correct.

The angular frequency of a torsional pendulum is $\omega$ rad/s. If the moment of inertia of the object is I, the torsional constant of the wire is related to the rotational kinetic energy of the disc, if the disc was rotating with an angular velocity $\omega$ is

  1. k= 2 KE

  2. k= KE

  3. k= 4 KE

  4. k= KE/2


Correct Option: A
Explanation:

we know that $T=2 \pi \sqrt{I/k}$. Substituting the values given, we get, $k= I \omega^2 =2 \times $ kinetic energy

The correct option is (a)

A small sphere is suspended by a string from the ceiling of a car. If the car begins to move with a constant acceleration $a$, the inclination of the string with the vertical is:-

  1. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ in the direction of motion

  2. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ opposite to the direction of motion

  3. ${\tan ^{ - 1}}\left( 2 \right)$ in the direction of motion

  4. ${\tan ^{ - 1}}\left( 2 \right)$ opposite to the direction of motion


Correct Option: B

A string of a simple pendulum can bear maximum tension that is $1.5$ times the weight of the bob. What is the maximum angle with which it can oscillate?

  1. $\cos^{-1} (0)$

  2. $\cos^{-1} (0.75)$

  3. $\cos^{-1} (0.25)$

  4. $\cos^{-1} (\sqrt {3}/2)$


Correct Option: A

A simple pendulum with a metal bob has a time period $T$. Now the bob is immersed in a liquid which is non viscous. This time the time period is $4T$. The the ratio of densities of metal bpob and that of the liquid is

  1. $15:16$

  2. $16:15$

  3. $1:16$

  4. $16:1$


Correct Option: C

A pendulum clock keeping correct time is taken to high altitudes,

  1. it will keep correct time

  2. its length should be increased to keep correct time

  3. its length should be decreased to keep correct time

  4. it cannot keep correct time even if the length is


Correct Option: C

Restoring force on the bob of a simple pendulum of mass $100\ gm$ when its amplitude is ${ 1 }^{ 0 } $ is 

  1. $0.017\ N$

  2. $1.7\ N$

  3. $0.17\ N$

  4. $0.034\ N$


Correct Option: A

Simple pendulum of large length is made equal to the radius of earth. Its period of oscillation will be then?

  1. 83.5 minutes

  2. 59.8 minutes

  3. 42.3 minutes

    1. 15 minutes

Correct Option: A
Explanation:

The time period of simple pendulum is given by:

$T=2\pi \sqrt{\dfrac{L}{g}}$

When length of pendulum is equal to the radius of earth. $R=L=6371\,\,Km=6371\times {{10}^{3}}\,Km$

So, time is

$ T=2\pi \sqrt{\dfrac{6371\times {{10}^{3}}}{10}} $

$ T=2\pi \times 798.18 $

$ T=5012.604\,seconds $

$\therefore$ $ T=83.54\,minutes $

If the length of a clock pendulum increase by $0.2\%$ due to atmospheric temperature rise, then the loss in time of clock per day is 

  1. $86.4$s

  2. $43.2$s

  3. $72.5$s

  4. $32.5$s


Correct Option: A
Explanation:
$T=2\pi \sqrt{\dfrac{l}{g}}$

$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\dfrac{\Delta l}{l}\times 100 $

$\dfrac{\Delta T}{24 \times 3600}\times 100 = \dfrac{1}{2}\dfrac{0.2}{100}\times 100 $

$\Delta T=86.4s$

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are
attached at distance $'L/2'$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio m/M is close to :

  1. 0.175

  2. 0.375

  3. 0.575

  4. 0.775


Correct Option: B
Explanation:

Frequency of torsonal oscillations is given by 
$f = \dfrac{k}{\sqrt{I}}$
$f _1 = \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12}}}$
$f _2 =  \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12} + 2m \left(\dfrac{L}{2} \right)^2}}$
$f _2 = 0.8 f _1$
$\dfrac{m}{M} = 0.375$

Find the time period of oscillations of a torsional pendulum, if the torsional constant of the wire is K = 10$\pi^2$J/rad. The moment of inertia of rigid body is 10 Kg m$^2$ about the axis of rotation.

  1. 2 sec

  2. 4 sec

  3. 16 sec

  4. 8 sec


Correct Option: A
Explanation:

Time period of a torsional pendulum is given by
$T = 2 \pi \sqrt{\dfrac{I}{k}}$
$\Rightarrow T=2 \pi \sqrt{\dfrac{10}{10\pi^2}}=2 sec$

A clock which has a pendulum made of brass keep correct time at ${30^0}C$? How many seconds it will gain or lose in day if the temperature falls to ${0^0}C$.

  1. It will lose $23.32$ sec per day

  2. It will gain $23.32$ sec per day

  3. It will lose $50$ sec per day

  4. It will gain $50$ sec per day


Correct Option: A

A simple pendulum of length $1$m has a bob of mass $100$g. It is displaced through an angle of $60^o$ from the vertical and then released . Find out K.E. of bob when it passes through mean position.

  1. $0.12$J

  2. $0.24$J

  3. $0.36$J

  4. $0.55$J


Correct Option: D
Explanation:

Length of simple pendulum $=1m$

Mass $=1w\;gm=0.1kg$
It is displaced through as angle of $60$ for vertical 
Height of pendulum at starting position $=$ length $(1-ws\;60)$
                                                                   $=1\left( 1-0.5\right)$
                                                                   $=0.5m$
Potential energy $=mgh=0.1\times 10\times 0.5 =0.55$
when it is released and it reaches mean position its potential energy at starting point is converted to kinetic energy.
so K.E. of bob at mean position $=0.55.$
Hence, the answer is $0.55.$

A pendulum is formed by pivoting a long thin rod of length L and mass m about a point P on the rod which is a distance d above the center of the rod as shown. 
Now answer the following questions. 
1. The time period of this pendulum when d = L/2 will be

  1. $2\pi\sqrt { \dfrac { 2\ell }{ 3g }}$

  2. $2\pi\sqrt { \dfrac { 3\ell }{ 2g }}$

  3. $4\pi\sqrt { \dfrac { \ell }{ 3g }}$

  4. $\dfrac {2\pi} {3} \sqrt { \dfrac { 2\ell }{ g }}$


Correct Option: C

A uniform rope of mass M =0.1 kg and length L=10 m hangs from the ceiling. [$  g=10 m /{ s }^{ 2 } $]

  1. Speed of the transverse wave in the rope increases linearly from bottom to the top with distance.

  2. Speed of the transverse wave in the rope decreases linearly from bottom to the top with distance.

  3. Speed of the transverse wave in the rope remains constant along the length of the rope

  4. Time taken by the transverse wave to travel the full length of the rope is 2 sec


Correct Option: D

The time period of a simple pendulum is 2.5 second. What will be it's total number of oscillations in 50 seconds ?

  1. 10

  2. 15

  3. 20

  4. None of these.


Correct Option: C

A simple pending of length l has a bob of mass m, with a charge q on it . A  vertical sheet of charge, with surface charge density $\sigma $ passes string makes an angle $\theta $ with the vertical , then 

  1. $\quad tan\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $

  2. $\quad tan\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $

  3. $\quad cot\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $

  4. $\quad cot\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $


Correct Option: A

The bob cf a simple pendulum is a spherical hollowe bal filled with water A pluyged hole near the bouthmol th oscilloting bob gets suddenly unplugged. During observation, till water is coming out, the time period of would 

  1. First increase and then decrease to the original value

  2. first decrease and then increase to the original value

  3. remain unchanged

  4. none of these


Correct Option: A

If the simple pendulum maximum kinetic energy of length 'I'  has maximum angular displacement $\theta $ then the maximum kinetic energy of the bob of mass 'm' is:

  1. $m g / ( 1 - \cos \theta )$

  2. $\operatorname { mg } l ( 1 - \sin \theta )$

  3. $m g / ( 1 + \cos \theta )$

  4. $m g l ( 1 + \sin \theta )$


Correct Option: A

Time period of a sample pendulum is T, time taken by it to complete 3/8 oscillations (in terms of distance traveled) is 

  1. $
    \cfrac { 3 T } { 12 }
    $

  2. $
    \cfrac { 3 T } { 8 }
    $

  3. $
    \cfrac { 7 T } { 12 }
    $

  4. $
    \cfrac { 5 T } { 12 }
    $


Correct Option: D

A metallic disc oscillates about an axis through its edge in it's own plane. The equivalent length of the disc as a pendulum is

  1. $r$

  2. $\dfrac r3$

  3. $\dfrac { r } { 2 }$

  4. $\dfrac { 3r } { 2 }$


Correct Option: D

The bob of a simple pendulum executes  $S H M$  in water with a period  $t,$  while the period of oscillation of the bob is  $t _{ 0 }$  in air. Neglecting the frictional force of water and given that the density of the bob is  $( 4 / 3 ) \times 1000 kg / { m } ^ { 3 }.$  What relationship between  $t$  and  $t _ { 0 }$  is true ?

  1. $t = t _ { 0 }$

  2. $t = 4 t _ { 0 }$

  3. $t = 2 t _ { 0 }$

  4. $t = t _ { 0 } / 2$


Correct Option: C

A simple pendulum is released when $\theta = \pi/6$. The time period of oscillation is

  1. $\displaystyle 2\pi\sqrt{\frac{l}{g}}$

  2. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{293}{288}\right)$

  3. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{288}{293}\right)$

  4. none of these


Correct Option: B
Explanation:

For large amplitudes, the time period is given by 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(1+\dfrac{\theta ^{2}}{16})$
Substitute $\theta =\dfrac{\pi }{6}$, we get answer as 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(\dfrac{293}{288})$
Option B is correct.

A pendulum suspended from the ceiling of an elevator at rest has time period ${ T } _{ 1 }$. When the elevator moves up with an acceleration 'a' its time period of oscillation becomes ${ T } _{ 2 }$ when the elevator moves down with an acceleration 'a', its period of oscillation become ${ T } _{ 3 }$ then

  1. ${ T } _{ 1 }=\sqrt { { T } _{ 2 }{ T } _{ 3 } } $

  2. ${ T } _{ 1 }=\sqrt { T _{ 2 }{ ^{ 2 }T _{ 3 } }^{ 2 } } $

  3. ${ T } _{ 1 }=\dfrac { \sqrt { 2 } { T } _{ 2 }{ T } _{ 3 } }{ \sqrt { {T _{ 2 }}^{ 2 }+{T _{ 3 } }^{ 2 } } } $

  4. ${ T } _{ 1 }=\dfrac { { T } _{ 2 }{ T } _{ 3 } }{ \sqrt { {T _{ 2 }}^{ 2 }+{T _{ 3 } }^{ 2 } } } $


Correct Option: D

The centripetal acceleration of the bob of a conical pendulem is......................

  1. $\dfrac{mg}{cos \theta}$

  2. $\dfrac{rg}{L}$

  3. $\dfrac{g}{L}$

  4. $\dfrac{g}{L cos \theta}$


Correct Option: A

The time period of a torsional pendulum is

  1. $T=\pi\sqrt{C/I}$

  2. $T=2\pi\sqrt{I/C}$

  3. $T=2\pi\sqrt{C/I}$

  4. $T=\pi\sqrt{I/C}$


Correct Option: B
Explanation:

In torsional pendulum, $\omega =\sqrt{\frac{C}{I}}$
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{C}}$
Option B is correct.

In a conical pendulum, when the bob moves in a horizontal circle of radius r, with uniform speed V, the string of length L describe a cone of semi-vertical angle $\theta$. The tension  in the string is given by 

  1. $T = \dfrac{mgl}{(L^2 - r^2)}$

  2. $ T = \dfrac{\sqrt {L^2 - r^2}}{mgl}$

  3. $ T = \dfrac{mgL}{\sqrt {L^2 - r^2}}$

  4. $ T = \dfrac{mgL}{(L^2 - r^2)^2}$


Correct Option: C

A seconds pendulum is attached to roof of car that is moving with acceleration $10{ m/s }^{ 2 }$ a straight road. Its frequency of oscillation 

  1. $\dfrac { 1 }{ 2 } Hz$

  2. $1Hz$

  3. $\dfrac { 1 }{ \sqrt { 2 } } Hz$

  4. ${ 2 }^{ -1/4 }Hz$


Correct Option: D

A simple pendulum in which the bob swings in a horizontal circle is called.

  1. Compound pendulum

  2. Horizontal pendulum

  3. Conical pendulum

  4. Gallitzin pendulum


Correct Option: C

The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\boxed { ? } $, is given by

  1. $2\pi \sqrt { \dfrac { L }{ gcos\alpha } } $

  2. $2\pi \sqrt { \dfrac { L }{ gsin\alpha } } $

  3. $2\pi \sqrt { \dfrac { L }{ g } } $

  4. $2\pi \sqrt { \dfrac { L }{ gtan\alpha } } $


Correct Option: A

A simple pendulum of length $40\ cm$ oscillates with an angular amplitude of $0.04\ rad$. Find the speed of the bob when the string makes $0.02 \ rad$ with the vertical. 

  1. $4.2\ cm/s$

  2. $3.4\ cm/s$

  3. $6.8\ cm/s$

  4. $13.6\ cm/s$


Correct Option: C

A metre  stick oscillates  as a compound pendulum  about a horizontal axis  through A Then 

  1. the length of an equivalent simple pendulum is 0.58 m

  2. the period of oscillation bout A and B is same

  3. The period of oscillation abut B is approximately 1.52 s

  4. the period of oscillation about A is approximately 2.45 s


Correct Option: C

A simple pendulum of length $40 cm$ oscillates with an angular amplitude of $0.04\ rad$. Find the angular acceleration when the bob is in momentary rest. Take $\displaystyle g=10: : m/s^{2}$.

  1. $\displaystyle 2: rad/s^{2}$

  2. $\displaystyle 3: rad/s^{2}$

  3. $\displaystyle 1: rad/s^{2}$

  4. $\displaystyle 4: rad/s^{2}$


Correct Option: C

The string of a simple pendulum in attached with the ceiling of a car moving on a straight horizontal raod with an acceleration $a=\dfrac {g}{\sqrt3}$, where $g$ is acceleration due to gravity near earth surface. The pendulum is made to oscillate at an angular amplitude of $30^o$. If the tension in the string is maximum when the string makes an angle $\theta$ with the vertical, then value of $\theta$ is 

  1. zero degree

  2. $30^o$

  3. $45^o$

  4. $60^o$


Correct Option: A

A disc of masses m and radius 2r are suspended through a fine wire of torsional constant K. The wire is attached to the centre of the plane of the disc and given torsional oscillations. If the disc is replaced by another disc of mass 4m and radius 2r, the ratio of the time period of oscillations are

  1. 4:1

  2. 1:4

  3. 1:1

  4. 2:1


Correct Option: D

A pendulum bob has a speed of $ 3 $ $ \mathrm{ms}^{-1} $ at  its lowest position. The pendulum is$ 0.5$ $ \mathrm{m}  $ long. The speed of the bob, when the length makes an angle of $ 60^{\circ}  $ to the vertical will be $ (g=10 $ $ \left(n s^{-1}\right) $

  1. $3$ $ m s^{-1} $

  2. $
    1 / 3 \mathrm{ms}^{-1}
    $

  3. $
    1 / 2 m s^{-1}
    $

  4. $
    2 m s^{-1}
    $


Correct Option: D
Explanation:
Apply energy conservation theorem$,$ 
energy at lowest position of Bob $=$ energy$ ,$ when Bob makes $60°$ to the vertical 
$1/2 mv^2 = 1/2 mv₁^2 + mgl(1 - cos60°)$
Here $v$ is speed at Lowest position $, v₁$ is speed $,$ when it makes $60°$ with vertical and $l$ is length of pendulum $.$
$[$Actually, height of Bob $,$ when it makes $60°$ with vertical $= l(1 - cos60°)] $
$∴ v^2 = v₁^2 + 2gl(1 - cos60°)$ 
$3^2 = v₁^2 + 2 × 10 × 0.5 (1 - 1/2)$ 
$9 = v₁^2 + 5$ 
$v₁^2 = 4 ⇒v₁ = 2m/s $
So$,$ speed of Bob $= 2m/s$
Hence,
option $(D)$ is correct answer.

Which of the following will change the time period as they are taken to moon?

  1. A simple pendulum

  2. A physical pendulum

  3. A torsional pendulum

  4. A spring-mass system


Correct Option: A,B
Explanation:

$(i)$ For simple pendulum $T = 2\pi\sqrt{L/g}$
$(ii)$ For physical pendulum $T = 2\pi\sqrt{I/mgL}$
So in both above case, time period is changed if they are taken to the moon.
$(iii)$ For torsional pendulum $T = 2\pi\sqrt{I/C}$
$(iv)$ For spring-mass system $T = 2\pi\sqrt{m/k}$

A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium positions, its time period of oscillation is:

  1. $ T = 2 \pi \sqrt{\dfrac{L}{g}}$

  2. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}+ \dfrac{v^4}{R^2}}}$

  3. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}+ \dfrac{v^2}{R}}}$

  4. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}- \dfrac{v^4}{R^2}}}$


Correct Option: B
Explanation:

The period of the pendulum, $T=2\pi \sqrt{\cfrac{L}{a}}$ where $a=$ $\text{resultant acceleration}\=\sqrt{g^2+{(\cfrac{V^2}{R})}^2}\quad\quad\quad\quad [\cfrac{V^2}{R}=\text{centripital acceleratiop }, g=\text{acceleration due to gravity}]$.

$\therefore T=2\pi\sqrt{\cfrac{L}{g^2+\cfrac{V^4}{R^2}}}$
Option B is the correct answer.


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