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

The error in the measurement of length of a simple pendulum is $0.1\%$ and error in the time period is $2 \%$. The possible maximum error in the quantity having dimensional formula $LT^{-2}$ is

  1. $1.1 \%$
  2. $2.2\%$
  3. $4.1\%$
  4. $6.1\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

Error $\%$ in time $=2\%=\dfrac{2}{100}=\dfrac{1}{50}=2\%$
Error $\%$ in length $=1\%$
Find $\%$ error in $\dfrac{1}{12}$
$\therefore$ as $\dfrac{\triangle A}{A}=\left( \dfrac{\triangle 2}{2}\right) +\left( \dfrac{\triangle Q}{Q}\right)$    if $A=\dfrac{t}{Q}$
Thus, 
$\Rightarrow \%$ error $=\left( \dfrac{\triangle L}{L}\right) +2\left( \dfrac{\triangle T}{T}\right)$
                   $=\left( 2\%\right) +2\left( .1\%\right)$
                   $=2\%+.2\%$
                   $=2.2\%$
Hence, the answer is $2.2\%.$

Multiple choice physics heat and temperature anomalous expansion of water thermal expansion in solids, liquids and gases thermal expansion

If the pendulum of a clock is made of a metal like steel or brass

  1. It's length increases in summer

  2. Time period of oscillation increases

  3. Both

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

If the pendulum of a clock is made of a metal like steel or brass,its length increases in summer due to increase in Temperature ,hence time period of oscillation increases and the clock goes slow.

Multiple choice physics heat and temperature anomalous expansion of water thermal expansion in solids, liquids and gases thermal expansion

A clock which keeps correct time at $20^{\small\circ}C$, is subjected to $40^{\small\circ}C$. If coefficient of linear expansion of the pendulum is $12\times10^{-6}$ per $^{\small\circ}C$. How much will it gain or loose in time?

  1. $10.3\space seconds/day$
  2. $20.6\space seconds/day$
  3. $5\space seconds/day$
  4. $20\space minutes/day$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\quad T = 2\pi\sqrt{\displaystyle\frac{l}{g}}$
$\quad \displaystyle\frac{\triangle T}{T} = 2\pi\times\displaystyle\frac{1}{2}(l/g)^{-1/2}\times\triangle l/l$
$\quad (\because T + \triangle T = 2\pi\sqrt{\displaystyle\frac{l+\triangle l}{g}})$
$\quad \therefore \displaystyle\frac{\triangle T}{T} = \displaystyle\frac{1}{2}\displaystyle\frac{\triangle l}{l} = \displaystyle\frac{1}{2}\propto\triangle\theta$
$\quad = \displaystyle\frac{1}{2}\times12\times10^{-6}\times(40-20) = 12\times10^{-5}$
$\quad \triangle T = T\times12\times10^{-5} = 24\times60\times60\times10^{-5} = 10.3\space s/day$

Multiple choice physics heat and temperature anomalous expansion of water thermal expansion in solids, liquids and gases thermal expansion

Why the error creeps into the time shown by the pendulum clock made of ordinary metal during winter and summer.

  1. In summer days are long and in winter it is small

  2. In summer, the length of the pendulum decreases, and in winter, it increases.

  3. In summer, the length of the pendulum increases, and in winter, it decreases.

  4. In summer days are small and in winters it is long
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time taken for the pendulum to complete one oscillation is called period $(T)$. $T$ increases with increase in length. We know that substances expand on heating and contract on cooling.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

A simple pendulum of mass m charged negatively to q coulomb oscillates with a time period T in a downward electric field E such that mg > qE. If the electric field is withdrawn, the new time period :

  1. $=$T
  2. $>$T
  3. $<$T

  4. any of the above three is possible
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Time period in the absence of electric field $T' = 2 \pi \displaystyle \sqrt{\frac{l}{g}}$

In the presence of electric field
$g _{eff}=(mg - qE) / m$
Therefore $T = 2 \pi \displaystyle \sqrt{\frac{l}{(mg - qE) / m}}$
$\displaystyle T'  = 2\pi \sqrt{\frac{l}{g}}$
or $T' < T$

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

Two pendulums of length $1.21m$ and $1.0m$ start vibrating. At some instant, the two are in the mean position in same phase. After how many vibrations of the longer pendulum, the two will be in phase?

  1. $10$
  2. $11$
  3. $20$
  4. $21$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a pendulum is T = 2*pi*sqrt(L/g). T1/T2 = sqrt(L1/L2) = sqrt(1.21/1.0) = 1.1 = 11/10. For them to be in phase, n1*T1 = n2*T2. n1/n2 = T2/T1 = 10/11. After 11 vibrations of the longer pendulum (T1), the shorter one (T2) will have completed 10 vibrations.

Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

A simple pendulum of length 4 m is taken to a height $R$ (radius of the earth) from the earth's surface.The time period of small oscillations of the pendulum is $(g _{surface}={\pi}^{2 } m{s}^{-2})$

  1. 2 s

  2. 4 s

  3. 8 s

  4. 16 s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The acceleration due to gravity of earth at a high 'h' 

${g} _{h} = {g} _{surface}\left( 1+\dfrac { h }{ R }  \right) ^{ -2 }$
g = acceleration due to surface gravity at earth surface
Now h = R (given)
so, ${g} _{h} = \dfrac { { g } _{ surface } }{ 4 } $
Now the time period of simple pendulum of length 4m at earth surface is 
${ T } _{ surface }=2\pi \sqrt { \dfrac { l }{ { g } _{ surface } }  } =2\pi \sqrt { \dfrac { 4 }{ 9.8 }  } \approx 4sec$
So time period at hight 'h = 2R' is 
$T=2\pi \sqrt { \dfrac { l }{ g _h }  } =2\pi \sqrt { \dfrac { 4\times 4 }{ { g } _{ surface } }  } =8sec$

Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

The oscillations of a pendulum about a vertical equilibrium position is an example of

  1. damped vibration

  2. free vibration

  3. forced vibration

  4. random vibration
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The oscillation of a pendulum about a vertical equilibrium position is an example of free vibration. As there no exciting force is present.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A pendulum with time of 1 s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in $s^{-1}$) is

  1. 2
  2. $\dfrac{1}{15} ln 3$
  3. $\dfrac{1}{2}$
  4. $\dfrac{1}{30} ln 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy E(t) = E0 * exp(-2kt). E(15) = 15, E0 = 45. 15 = 45 * exp(-2k * 15). 1/3 = exp(-30k). ln(1/3) = -30k. k = ln(3)/30.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

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

The clock loses time because its period is longer than the standard 2 seconds. The time lost in one day is (delta_T / T_actual) * 86400 seconds. With delta_T = 0.04s and T = 2s, the loss is (0.04/2.04) * 86400 approx 1694 seconds, which is about 28.2 minutes.