Tag: measurements and experimentation

Questions Related to measurements and experimentation

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

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

  1. 4πG/3gR

  2. 3πR/4gG

  3. 3g/4πRG     

  4. πRg/12G

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Correct Option: C
Explanation:

We know that

$g=\cfrac{GM}{R^2}$
Also, density $=mass\times volume$
$M=density\times volume\M=P\times\cfrac{4\pi R^3}{3R^2}=P\times\cfrac{4\pi R}{3}$
Put value of m in $g=\cfrac{GM}{R^2}\P=\cfrac{3g}{4\pi RG}$

Let the time period of a seconds pendulum is $2.5\ s.$ Tell by how much time will the clock behind in $10\ hrs.$

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

  1. $2.5\ hr$

  2. $2\ hr$

  3. $1.5\ hr$

  4. $None$

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Correct Option: C

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will :

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

  1. Be less

  2. Be more

  3. Remain unchanged

  4. None of these

Show answer
Correct Option: C
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. 

If the length of a seconds pendulum is increased by $2$% then what is loss and gain in a day?

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

  1. losses $764 \ s$

  2. losses $924 \ s$

  3. gains $236 \ s$

  4. losses $864 \ s$

  5. gains $346 \ s$

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Correct Option: D
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$=(0.01\times24\times3600)seconds\=864seconds$

If the length of second's pendulum is increased by $2\%$, how many second will it lose per day?

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

  1. $923$ s

  2. $3727$ s

  3. $3642$ s

  4. $864$ s

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Correct Option: D

The different equation of simple harmonic motion for a seconds pendulum is:

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

  1. $\dfrac{d^2 x}{dt^2} + x = 0$

  2. $\dfrac{d^2 x}{dt^2} + \pi x = 0$

  3. $\dfrac{d^2 x}{dt^2} + 4 \pi x = 0$

  4. $\dfrac{d^2 x}{dt^2} + \pi^2 x = 0$

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Correct Option: A

A simple pendulum with a bob of mass m swings with an angular amplitude of ${ 60 }^{ 0 }$, when its angular displacement is ${ 30 }^{ 0 }$, the tension of string would be 

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

  1. $3\sqrt { 3 } mg$

  2. $\frac { 1 }{ 2 } mg(2\sqrt { 3 } -1)$

  3. $\frac { 1 }{ 2 } mg(3\sqrt { 3 } +2)$

  4. $\frac { 1 }{ 2 } mg(3-\sqrt { 2 } )$

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Correct Option: B

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:

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

  1. $\sqrt { 2 } s$

  2. $2\sqrt { 2 } s$

  3. $2s$

  4. $\dfrac { 1 }{ \sqrt { 2 } } s$

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Correct Option: B
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) $
  $=2\sqrt { 2 } s$

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.

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

  1. True

  2. False

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Correct Option: B
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.

A second's pendulum can be used as a timing device

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

The second's pendulum has a definite time period of 2 seconds. We can calculate the time by counting its no.of oscillation. So, it can be used as a time measuring equipment.