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 maths the trapezium rule approximation errors and approximations the need for approximation

The period of oscillation $T$ of a pendulum of length $l$ at a place of acceleration due to gravity $g$ is given by $T=2\pi \sqrt {\dfrac {l}{g}}$. If the calculated length is $0.992$ times the actual length and if the value assumed for $g$ is $1.002$ times its actual value, the relative error in the computed value of $T$ is

  1. $0.005$
  2. $-0.005$
  3. $0.003$
  4. $-0.003$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Relative error will calculate by
$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{\Delta L}{L}-\dfrac{\Delta g}{g}\right]$

$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{0.992L}{L}-\dfrac{1.002 g}{g}\right]$

$\Delta T=-0.005T$

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

A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a nail fixed at its circumference and perpendicular to plane of ring. The time period will be

  1. 1/4 sec

  2. 1/2 sec

  3. 2sec

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

For a physical pendulum (ring oscillating about a nail on its circumference), the time period T = 2*pi * sqrt(I / mgd). Here, I = I_cm + md^2 = (1/2)mr^2 + mr^2 = (3/2)mr^2. With d = r, T = 2*pi * sqrt((3/2)mr^2 / mgr) = 2*pi * sqrt(3r / 2g). With diameter 1m, r = 0.5m. T = 2*pi * sqrt(1.5 / 9.8) approx 2 seconds.

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

The time taken to complete $20$ oscillations by a seconds pendulum is: 

  1. $20s$
  2. $50s$
  3. $40s$
  4. $5s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that the time period of a seconds pendulum is $T=2$ sec. One second for a swing in one direction and one second for the return swing. 

Thus, time taken to complete one oscillation is $2$ sec.
Hence, time taken to complete 20 oscillations is $2\times 20=40$ sec.

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 on the surface of the earth is equal to 99.49 cm. True or false.

  1. True

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

The time period of seconds pendulum T = 2 seconds, acceleration due to gravity at earth g= 980 $\dfrac { cm }{ { s }^{ 2 } } $,It '$l$' is the length of pendulum,

$l=\dfrac { { T }^{ 2 }g }{ 4{ \pi  }^{ 2 } } \ \Rightarrow l=\dfrac { 4\times 980 }{ 4\times \left( \dfrac { 22 }{ 7 }  \right) ^{ 2 } } =\dfrac { 4\times 980\times 49 }{ 4\times 489 } =99.49$

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 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 ,     $l _{m}=1/6m$