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 evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

Time period of a simple pendulum inside a satellite orbiting earth is

  1. Zero
  2. $\infty$
  3. $T$
  4. $2T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As we know, $T\alpha \dfrac { 1 }{ \sqrt { g }  } $

On a artificial satellite, orbiting the earth, neg gravity is zero. As,
$\Rightarrow \quad g=0\Rightarrow T\rightarrow \infty $
$\Rightarrow$  Option B is the correct answer.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

The time period of a vibrating body is $0.01\;sec$. Its frequency will be :

  1. $1.0\;s^{-1}$
  2. $10.0\;s^{-1}$
  3. $100.0\;s^{-1}$
  4. $1000.0\;s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


Time period $=0.01\,sec$

We have to find the frequency

We have the relation,

$Frequency=\frac{1}{Time\,Period}$

That is,

$Frequency=\frac{1}{0.01}=100\,s^{-1}$

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

The average speed of the to bob of a seconds pendulum is $2 {ms}^{-1}$. Determine the frequency of oscillation.

  1. 0.5 Hz

  2. 1.5 Hz

  3. 0.9 Hz

  4. 2 Hz
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a pendulum, average speed = total distance / total time. In one oscillation, the bob travels 4A distance in time T. Average speed = 4A / T = 4Af. Given average speed = 2 m/s and length of a seconds pendulum is 1m (T=2s), we find the frequency.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Which of the following is not an oscillatory motion?

  1. The tuning fork

  2. The stretched string

  3. The motion of the swing

  4. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Oscillation is a motion in which a body moves back and forth repeatedly about a mean position. Since in all the first 3 options, the motion is oscillatory, the correct option is 'None of these'.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

What is the number of degrees of freedom of an oscillating simple pendulum?

  1. more than three

  2. 3

  3. 2

  4. 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There are 2 degrees of freedom, 1 translational along which the pendulum bob moves, and second rotational - the hinge about which it forms an arc motion.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

An object swinging on the end of a string forms a simple pendulum. Some students (and some texts) often cite the simple pendulum's motion as an example of SHM. That is not quite accurate because the motion is really

  1. approximately SHM only for small amplitudes

  2. exactly SHM only for amplitudes that are smaller than a certain value

  3. approximately SHM for all amplitudes.

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An object performing SHM moves along a straight path.

For large amplitudes, a pendulum moves in a curved path.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Which of  the  following regarding oscillatory motion is true?

  1. Motion of the earth is periodic but not oscillatory because it is not to and fro.

  2. Quivering of the string of the musical instrument is an example of oscillatory motion

  3. Motion of the earth is periodic and oscillatory motion because it is not to and fro.

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Oscillator motion is to and fro motion about a mean position. Earth motion is not a to and fro motion here, hence it is not an oscillatory motion. But as earth motion is repeated in a regular interval of time, its motion is periodic.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

In an electronic watch, the component corresponding to the pendulum of a pendulum clock is a__?

  1. Diode

  2. Transistor

  3. Crystal oscillator

  4. Balance wheel
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electronic watches use a quartz crystal oscillator to provide a stable frequency for timekeeping, replacing the mechanical pendulum or balance wheel.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The period of oscillation of a simple pendulum is Given by $ T=2\pi \sqrt { \frac { \ell  }{ g }  }$  where  $\ell$ is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measured by a stop watch of least count 0.1 s. The percentage error in g is:-

  1. 0.1 %

  2. 1 %

  3. 0.2 %

  4. 0.8 %
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

T = 2*pi*sqrt(L/g) implies g = 4*pi^2*L/T^2. Relative error dg/g = dL/L + 2*dT/T. dL = 0.1 cm, L = 100 cm, dL/L = 0.001. dT = 0.1/100 = 0.001 s, T = 2 s, dT/T = 0.0005. dg/g = 0.001 + 2(0.0005) = 0.002 or 0.2%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The time period of oscillation of simple pendulum given by $T = 2\pi\sqrt{\frac{L}{g}}$ where $L= (200 \pm 0.1) cm$. The time period, T =4s and the time of 100 oscillations is measured using a stopwatch of least count 0.1 s. The percentage error in g is

  1. $0.1$%
  2. $1.5$%
  3. $2$%
  4. $4$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

g = 4*pi^2*L/T^2. dg/g = dL/L + 2*dT/T. dL = 0.1, L = 200, dL/L = 0.0005. dT = 0.1/100 = 0.001, T = 4, dT/T = 0.00025. dg/g = 0.0005 + 2(0.00025) = 0.001 or 0.1%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His per-centre error in the n=measurement of g by the relation $g={ 4\pi  }^{ 2 }\left( I/T^{ 2 } \right) $ will be 

  1. 2%

  2. 4%

  3. 7%

  4. 10%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given g = 4 * pi^2 * (L / T^2). The relative error is dg/g = dL/L + 2 * dT/T. Using magnitudes: 1% + 2 * 3% = 1% + 6% = 7%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The length of a pendulum is measured as $1.01$ m and time for $30$ oscillations is measured as $1$ minute $3$ seconds. Error in length is $0.01$ m and error in time is $3$ seconds. The percentage error in the measurement of acceleration due to gravity is:

  1. $1$
  2. $5$
  3. $10$
  4. $15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$\implies g=\dfrac{4\pi^2 l}{T^2}$

Thus $\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}$

Percentage error in measurement of: 
$g=\dfrac{\Delta g}{g}\times 100=(\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T})\times 100$
   $=(\dfrac{0.01}{1.01}+2\dfrac{3}{63})\times 100$% $=10$%

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

A students performs an experiment to determine the acceleration due to gravity (g) at a place using a simple pendulum. The length of the pendulum is 60 cm and the total time for 30 oscillations is 100s. What is maximum percentage error for the measurement g ? Given, least count for time $=0.1 s$ and least count for length $=0.1 cm$. 

  1. $0.26 \%$
  2. $0.3 \%$
  3. $0.36 \%$
  4. $3.6 \%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a pendulum, the time period is: $T=2\pi\sqrt{\dfrac{l}{g}}$ 
$\Rightarrow g=\dfrac{4\pi^2 l}{T^2}=\dfrac{4\pi^2 ln^2}{t^2}$

Take $ln$ and then differentiate:
$\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta t}{t}$   (as n is constant so its derivative will be zero)

The % error in g $=\dfrac{\Delta g}{g}\times 100=\dfrac{\Delta l}{l}\times 100+2\dfrac{\Delta t}{t}\times 100=\dfrac{0.1}{60}\times 100+2\dfrac{0.1}{100}\times 100=0.36 \%$

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is:

  1. $3%$
  2. $4%$
  3. $6%$
  4. $5%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

g = 4 * pi^2 * L / T^2. Relative error dg/g = dL/L + 2 * dT/T. Given dL/L = 1% and dT/T = 2%, dg/g = 1% + 2 * 2% = 5%.