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 free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

The phenomenon in which the amplitude of oscillation of a pendulum decreases gradually is called

  1. decay period of oscillation

  2. damping

  3. building up of oscillation

  4. maintained oscillation
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Whenever there is a damping force, it will slow down the motion of a pendulum, and ultimately it will make the pendulum stop. This phenomenon is called damping.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

The oscillations of a pendulum slow down due to :

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

  2. the force exerted by air only

  3. the forces exerted by friction at the support

  4. they never slow down
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The pendulum on its motion has friction particle. As a result of this frictional force slows down.

Multiple choice resonance oscillations physics

Which of the following is an example of mechanical resonance?

  1. A child on a swing.

  2. A pendulum.

  3. A tuning fork.

  4. Nuclear magnetic resonance
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$Answer:-$ A,B

Mechanical resonance is the tendency of a mechanical system to respond at greater amplitude when the frequency of its oscillations matches the system's natural frequency of vibration (its resonance frequency or resonant frequency) than it does at other frequencies. It may cause violent swaying motions and even catastrophic failure in improperly constructed structures including bridges, buildings and airplanes—a phenomenon known as resonance disaster.

Various examples of mechanical resonance include:-

  • Most clocls keep time by mechanical resonance in a balance wheel, pendulum, or quartz crystal.
  • The resonance of the basilar membranein the ear.
  • Making a child's swing swing higher by pushing it at each swing.
  • A wineglass breaking when someone sings a loud note at exactly the right pitch.

Multiple choice resonance oscillations physics

Which of the following shows mechanical resonance?

  1. Balance wheel

  2. Pendulum

  3. Quartz crystal

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Answer:-$ D

Mechanical resonance is the tendency of a mechanicalsystem to respond at greater amplitude when the frequency of its oscillations matches the system's natural frequency of vibration (its resonance frequency or resonant frequency) than it does at other frequencies.
examples: Most clocks keep time by mechanical resonance in a balance wheel, pendulum, or quartz crystal.

Multiple choice maths direct proportion and inverse proportion inverse proportion rule of three types of proportions

The length of a pendulum varies inversely as the square of the number of beats it makes per minute. If a pendulum, $65$ cm long, makes $27$ beats per minute, then the length of the pendulum that makes $24$ beats per minutes is 

  1. $91$ cm
  2. $85$ cm
  3. $81$ cm
  4. $71$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Length $\alpha \cfrac{1}{(\text {No. of beats} )^2}$
$ \Rightarrow L = \cfrac{k}{\text {(beat)}^2}$, where $k$ is constant.
$\Rightarrow 65 = \cfrac{k}{(27)^2}$
$\Rightarrow k = 65 \times (27)^2$ 
Also, $k=L\times(24)^2$
$\Rightarrow 65 \times (27)^2= L \times(24)^2$
$\Rightarrow L = 81 cm $(approx)
Multiple choice physics types of energy law of conservation of energy the law of conservation of energy work, energy and machines

When a pendulum oscillates, it comes to rest after sometime because :

  1. energy lost by pendulum to overcome friction is gained by pendulum

  2. energy lost by pendulum to overcome its speed is gained by surrounding

  3. energy lost by pendulum to overcome friction is gained by surrounding

  4. energy lost by pendulum to overcome friction is gained by surrounding and pendulum
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a pendulum oscillates, it comes to rest after sometime because energy lost by pendulum to overcome friction is gained by surrounding. Hence total energy of pendulum and surrounding system remains conserved.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be  

  1. $\pi \times { \left( \dfrac { (R-r)1.4 }{ g } \right) } $
  2. $2\pi \times { \left( \dfrac { R-r }{ g } \right) } $
  3. $\sqrt [ 2\pi ]{ \left( \dfrac { r\quad R }{ g } \right) } $
  4. ${ \left( \dfrac { R }{ g\quad r } \right) } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a sphere of radius r rolling/sliding in a concave mirror of radius R, the effective length of the pendulum is (R-r). The time period for a simple pendulum is T = 2*pi * sqrt(L/g). Substituting L = R-r, we get T = 2*pi * sqrt((R-r)/g).

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

Find the time period of small oscillations of the following systems. 

  1. A metre stick suspended through the 20 cm mark.

  2. A ring of mass m and radius r suspended through a point on its perphery.

  3. A uniform square plate of edge a suspended through a corner.

  4. A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A physical pendulum's time period is T = 2*pi*sqrt(I/mgd). Option A describes a physical pendulum where the moment of inertia and distance from the center of mass can be calculated to find the period.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A student measures the time period of oscillation of a simple pendulum. He uses the data to estimate the acceleration due to gravity 9g) at that place. If the maximum percentage error in measurement of length pendulum and that in time are $ e _{1} $ and $ e _{2} $ respectively then percentage error estimation of ''g'' is :

  1. $

    e _{1}+2 e _{2}

    $
  2. $

    2 e 1+e 2

    $
  3. $

    e 1+e _{2}

    $
  4. $

    e 1-e _{2}

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a simple pendulum, T = 2*pi*sqrt(l/g), so g = 4*pi^2*l / T^2. The relative error is dg/g = dl/l + 2*dT/T. Thus, the percentage error is e1 + 2*e2.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The angular frequency of the damped oscillator is given by $\omega =\sqrt { \left( \dfrac { k }{ m } -\dfrac { { r }^{ 2 } }{ 4{ m }^{ 2 } }  \right)  }$ , where k is the spring constant, $m$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\dfrac { { r }^{ 2 } }{ mk }$ is $80$%, the change in time period compared to the undamped oscillator is approximately as follows:

  1. Decreases by $1$%
  2. Increases by $8$%
  3. Increases by $1$%
  4. Decreases by $8$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The damped frequency is omega' = sqrt(omega0^2 - gamma^2). The time period T' = 2*pi/omega'. For small damping, T' approx T(1 + gamma^2 / (2*omega0^2)). With r^2/mk = 0.8, the change is small and negative/positive depending on the exact definition.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The period of oscillation of a simple pendulum of constant length is independent of

  1. size of the bob

  2. shape of the bob

  3. mass of bob

  4. all of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$T=2 \pi \sqrt{\dfrac{L}{g}}$               (where L= length of sting)
From above equation, Time period only depend on the length of the string and g.

Option d 

Multiple choice physics units and measurement: error analysis accuracy of measurement accuracy and precision accuracy, precision and uncertainty in measurement

The time of 25 oscillations of a simple pendulum is measured to be $50.0 s$ by a watch of least count $0.1 s$. The percentage error in time is

  1. $0.2 \%$
  2. $0.02 \%$
  3. $0.002 \%$
  4. $2 \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The total time is 50.0 s with a least count of 0.1 s. The absolute error is 0.1 s. Percentage error = (absolute error / total time) * 100 = (0.1 / 50.0) * 100 = 0.2%.