Physics

Oscillations and Periodic Motion

173 Questions

Oscillations and periodic motion describe the movement of objects repeating their paths in regular intervals. Key concepts include simple pendulums, kinetic energy variations, and mechanical resonance. This physics topic is vital for various competitive exams.

Simple pendulumTime period calculationsKinetic energy in SHMMechanical resonanceDamped oscillations

Oscillations and Periodic Motion Questions

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will :

  1. Be less

  2. Be more

  3. Remain unchanged

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time period of simple pendulum id given by

$T=2\pi \sqrt{\dfrac{l}{g}}$
where, $l=$ length of simple pendulum
$g=$ acceleration due to gravity
$T=$ Time period
The time period of simple pendulum is independent of the mass of bob, the time period remains unchanged,when mass of bob will change.
The correct option is C. 

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

If the length of a seconds pendulum is increased by $2$% then what is loss and gain in a day?

  1. losses $764 \ s$
  2. losses $924 \ s$
  3. gains $236 \ s$
  4. losses $864 \ s$
  5. gains $346 \ s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$T _0=2\pi\sqrt{\cfrac{l}{g}}\T^1=2\pi\sqrt{\cfrac{l+l\times2/100}{g}}\ \cfrac{T _0}{T^1}=\cfrac{\sqrt{100}}{\sqrt{102}}\ T^1=\cfrac{\sqrt{102}}{\sqrt{100}}T _0\T^1=1.0099T _0\approx  1.01T _0\Loss=(1.01-1)T _0=0.01T _0$

In one second, it looses $0.01sec$
$\Rightarrow$ Total time loose in one day$=(0.01\times24\times3600)seconds\=864seconds$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The different equation of simple harmonic motion for a seconds pendulum is:

  1. $\dfrac{d^2 x}{dt^2} + x = 0$
  2. $\dfrac{d^2 x}{dt^2} + \pi x = 0$
  3. $\dfrac{d^2 x}{dt^2} + 4 \pi x = 0$
  4. $\dfrac{d^2 x}{dt^2} + \pi^2 x = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The differential equation for simple harmonic motion is d^2x/dt^2 + omega^2 x = 0. For a seconds pendulum, the time period T = 2 seconds, so angular frequency omega = 2pi/T = 2pi/2 = pi. Substituting omega^2 into the equation gives d^2x/dt^2 + pi^2 x = 0.

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The simple pendulum acts as second's pendulum on earth. Its time on a planet, whose mass and diameter are twice that of earth is:

  1. $\sqrt { 2 } s$
  2. $2\sqrt { 2 } s$
  3. $2s$
  4. $\dfrac { 1 }{ \sqrt { 2 } } s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time period of second's pendulum is two second.
Second's pendulum is that simple pendulum whose time period of vibration is two seconds. The bob of such pendulum while oscillating passes through the mean position after every one second.
Noe,
Time period of simple pendulum is given by
$T=2\pi \sqrt { \left( \dfrac { l }{ g }  \right)  } $
or  $T\propto \dfrac { 1 }{ \sqrt { g }  } $             ......(i)
but  $g=\dfrac { GM }{ { R }^{ 2 } } $      (on earth)
and  ${ g }^{ \prime  }=\dfrac { G\left( 2M \right)  }{ 4{ R }^{ 2 } } $     (on planet)
$=\dfrac { 1 }{ 2 } \dfrac { GM }{ { R }^{ 2 } } =\dfrac { g }{ 2 } $
Equation (i) gives
$\dfrac { { T }^{ \prime  } }{ T } =\dfrac { \sqrt { g }  }{ \sqrt { { g }^{ \prime  } }  } =\sqrt { 2 } $
or  ${ T }^{ \prime  }=\sqrt { 2 } T$
  $=\sqrt { 2 } \times 2              \left( T=2s \right) $
  $=2\sqrt { 2 } s$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The length of a second's pendulum at a place where g = 9.8m/s $\displaystyle ^{2}$ is 90.2 cm. State whether true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time period of pendulum is:

$T =2\pi \sqrt [  ]{ \cfrac { l }{ g }  } $
$l=\cfrac { T^{ 2 }g }{ 4\pi ^{ 2 } } $
$l=\cfrac { 4\times 9.8 }{ 4\times \pi ^{ 2 } } $
$l=0.993m=99.3m$
$l$= length of pendulum 
$g$= $9.8m/s$
$T$ = Time period of seconds pendulum $=2s$
So, our given statement is false.

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The length of a second pendulum at the surface of earth is $1\ m$. The length of second pendulum at the surface of moon, where $g$ is $\dfrac{1}{6} th$ that of earth's surface.

  1. $\dfrac{1}{6} m$
  2. $6 m$
  3. $\dfrac{1}{36}m$
  4. $36 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that the time period of a second pendulum is $2s$ .

At earth ,
                $T=2\pi \sqrt{l _{e}/g _{e}}$
                $2=2\pi\sqrt{l _{e}/g _{e}}$
or             $g _{e}=\pi^{2}l _{e}$  ...............................eq1

At moon ,
                $T=2\pi \sqrt{l _{m}/g _{m}}$
or             $2=2\pi \sqrt{l _{m}/(g _{e}/6)}$  , given   $g _{m}=g _{e}/6$
or             $2=2\pi\sqrt{6l _{m}/\pi^{2}l _{e}}$   ,   putting the value of $g _{m}$ from  eq1
or             $l _{m}=l _{e}/6$
Now , given  $l _{e}=1m$

Hence ,     $l _{m}=1/6m$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The length of the simple pendulum which ticks seconds is:

  1. $0.5$m
  2. $1$m
  3. $1.5$m
  4. $2$m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a simple pendulum is
$T = 2 \pi \sqrt{\dfrac{L}{g}}$
where L is the length of the pendulum.
or $ L = \dfrac{gT^2}{4 \pi^2}$
The time period of the simple pendulum which ticks seconds is $2$s.
$\therefore T = 2s$
Substituting in (i), we get


$L = \dfrac{(9.8 m s^{-2})(2s)^2}{4 \times (3.14)^2} = 1m$

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

A second's pendulum is mounted in a rocket. Its period of oscillation will decrease when the rocket is:

  1. moving up with uniform velocity

  2. moving up with uniform acceleration

  3. moving down with uniform acceleration

  4. moving around the earth in a geostationary orbit
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Correct answer= B

As the rocket accelerates upwards, pseudo force acts in the opposite direction of propagation.
=> Pseudo force acts in downward direction and gets added up to gravitational force.
=> Effective gravity= gravitational force+ pseudo force
                                >Gravitational force
=>             g'        >         g       where g' = effective gravity
Since time period of oscillation of pendulum= √(L/g)
       where L= length of the pendulum
                   g= gravitational force acting on the pendulum
=> In this situation,
          time period of oscillation of pendulum=√(L/g')
Since g' > g
=>     √(L/g')      <     √(L/g)
=>  Time period of oscillation decreases

Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as

  1. simple pendulum

  2. torsional pendulum

  3. compound pendulum

  4. seconds pendulum
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a rigid body is suspended vertically, and it oscillates with a small amplitude under the action of the force of gravity, the body is known as compound pendulum. Thus the periodic time of a compound pendulum is minimum when the distance between the point of suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.

The correct option is (c)

Multiple choice physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

 The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\ cm$ to $8\ cm$ In $40$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$, the time In which amplitude of this pendulum will reduce from $10\ cm$ to $5\ cm$ in carbondioxide will be close to (in $5=1.601, \ln { 2 }  2=0.693$)

  1. $231\ s$
  2. $208\ s$
  3. $161\ s$
  4. $142\ s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The amplitude decay of a pendulum in a viscous medium follows the equation A = A_0 * exp(-bt/2m). The damping constant b is proportional to the viscosity eta. Since the ratio of viscosities is 1.3, the decay constant in CO2 is 1.3 times that in air. By comparing the time taken to reach half amplitude, the result is calculated as 161 seconds.

Multiple choice physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a constant rate. Which of the following statements describes the variation of the time period (T) of the pendulum as the water flows out?

  1. T decreases first and then increases.

  2. T increases first and then decreases.

  3. T increases throughout.

  4. T does not change.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
First distance of comfrom suspension point will increase then decrease.