Physics

Oscillations and Periodic Motion

162 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 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.

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

There is a clock which gives correct time at $20^o$C is subjected to $40^o$C. The coefficient of linear expansion of the pendulum is $12\times 10^{-6}$ per $^oC$, how much is gain or loss in time?

  1. $10.3$ sec/day
  2. $19$ sec/day
  3. $5.5$ sec/day
  4. $6.8$ sec/day
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fractional change in time period is given by delta T / T = 1/2 * alpha * delta theta. With alpha = 12*10^-6 and delta theta = 20 degrees, the fractional change is 1.2*10^-4. Multiplying by the number of seconds in a day (86400), we get approximately 10.36 seconds.

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

Find the length of a simple pendulum such that its time period is $2\ s$.

  1. $99.4\ cm$
  2. $89.4\ cm$
  3. $79.4\ cm$
  4. $109.4\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$


$T^2\, =\, 4\pi ^2\, \displaystyle \frac {L}{g}$
$\Rightarrow\, 2^2\, =\, 4\, \times\, 3.14\, \times\, 3.14\, \times\, \displaystyle \frac {L}{9.8}$
$\Rightarrow\, L\, =\, \displaystyle \frac {4\, \times\, 9.8}{4\, \times\, 3.14\, \times\, 3.14}\, m\, =\, 0.994\, m\, =\, 99.4\, cm$

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

A desktop toy pendulum swings back and forth once every $1.0 s$. How long is this pendulum?

  1. $0.25\, m$
  2. $0.50\, m$
  3. $0.15\, m$
  4. $0.30\, m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)


Putting $T = 1$ in eqn. (1), 


We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

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

You are designing a pendulum clock to have a period of $1.0\ s$. How long should the pendulum be ?

  1. $0.25\ m$
  2. $0.50\ m$
  3. $0.25\ cm$
  4. $0.25\ mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)

Putting $T = 1$ in eqn. (1), 

We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

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

Two pendulums of lengths 121 cm and 100 cm start vibrating at the same instant. They are in the mean position and in the same phase. After how many vibrations of the shorter pendulum, the two will be in the same phase in the mean position? 

  1. 10 vibrations

  2. 11 vibrations

  3. 21 vibrations

  4. 20 vibrations
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two pendulums of length $121cm$ and $100cm$.


Let,
$L _1=121cm=\dfrac{121}{100}=1.21m$

$L _2=100cm=\dfrac{100}{100}=1m$

We have to find the vibrations made by the shorter pendulum, such that both will be in same phase from the reaction,

$T _1=longer\,pendulum$


$T _2=shorter\,pendulum$


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

$T\propto \sqrt{L}$

$\dfrac{T _1}{T _2} \propto \sqrt{{L _1}{L _2}}$

$\dfrac{T _1}{T _2}\propto \sqrt{\dfrac{1.21}{1}}$

$\dfrac{T _1}{T _2}=\dfrac{1.1}{1}$

$10T _1=11T _2$

$10$ vibrations of longer pendulum= $11$ vibrations of shorter pendulum

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

A cylindrical block of wood $(density=650 kg m^{-3})$, of base area $30 cm^2$ and height $54 cm$, floats in a liquid of density $900 kg$ $m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly)

  1. 52 cm

  2. 26 cm

  3. 39 cm

  4. 65 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As block is floating it's weight should be equal to buoyancy force 
$\rho _{wood} V _{cylinder} g=\rho _{liquid} V _{displaced} g$
$V _{displaced} =\frac{\rho _{wood} V _{cylinder} }{\rho _{liquid}} $
$V _{displaced} =\frac{\rho _{wood} V _{cylinder} }{\rho _{liquid}} $...(i)
After displacing by small distance x, the net force on cylinder will be
$F=Buoyancy-w=\rho _{liquid}( V _{displaced} +A _{cylinder}\Delta x) g- \rho _{wood} V _{cylinder} g$
$F=\rho _{liquid}  V _{displaced} g+\rho _{liquid} A _{cylinder}\Delta x g-\rho _{wood} V _{wood} g$ the net force on cylinder becomes 
from equation (i) $\rho _{wood} V _{cylinder} g=\rho _{liquid} V _{displaced} g$
$F=\rho _{liquid} A _{cylinder}\Delta x g$
$ma=\rho _{liquid} A _{cylinder}\Delta x g$
$a=\frac{\rho _{liquid} A _{cylinder}}{m}\Delta x $
$a = \omega^2 \Delta x $
$\omega^2 =\frac{\rho _{liquid} A _{cylinder}}{\rho _{cylinder} V _{cylinder}}$
$\omega^2 =\frac{\rho _{liquid} A _{cylinder}}{\rho _{wood} A _{cylinder} h _{cylinder}}$
$\omega^2 =\frac{\rho _{liquid}}{\rho _{wood} h _{cylinder}}$
$\omega^2 =\frac{900}{650\times .54 }$
This should be equal to angular frequency of simple pendulum
$ \omega=\sqrt{\frac{g}{l}}$
$\sqrt{\frac{900}{650\times .54}}=\sqrt{\frac{g}{l}}$
$l=g\frac{650\times .54}{900}$
$l=.06\times 65$
$=39 cm$

Multiple choice uniform magnetic field lines of earth magnetism physics

The time period of a thin magnet is 4 s. If it is divided into two equal halves, then the time period of each part will be:

  1. 4s

  2. 1s

  3. 2s

  4. 8s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In case of vibration magnetometer when a magnet is cut n equal parts by cutting normal to its length. Then the time period of each part of magnet will be 
$T'=\frac{T}{n}$       ...(i)   (here, $T=4s, n = 2$)
Now, Putting the given values in Eq. (i), we get
$T'=\frac{4}{2}=2s$