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

You are designing a pendulum clock to have a period of $1.0\ s$. How long should the pendulum be ?

  1. $0.25\ m$
  2. $0.50\ m$
  3. $0.25\ cm$
  4. $0.25\ mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)

Putting $T = 1$ in eqn. (1), 

We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

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

Two pendulums of lengths 121 cm and 100 cm start vibrating at the same instant. They are in the mean position and in the same phase. After how many vibrations of the shorter pendulum, the two will be in the same phase in the mean position? 

  1. 10 vibrations

  2. 11 vibrations

  3. 21 vibrations

  4. 20 vibrations
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two pendulums of length $121cm$ and $100cm$.


Let,
$L _1=121cm=\dfrac{121}{100}=1.21m$

$L _2=100cm=\dfrac{100}{100}=1m$

We have to find the vibrations made by the shorter pendulum, such that both will be in same phase from the reaction,

$T _1=longer\,pendulum$


$T _2=shorter\,pendulum$


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

$T\propto \sqrt{L}$

$\dfrac{T _1}{T _2} \propto \sqrt{{L _1}{L _2}}$

$\dfrac{T _1}{T _2}\propto \sqrt{\dfrac{1.21}{1}}$

$\dfrac{T _1}{T _2}=\dfrac{1.1}{1}$

$10T _1=11T _2$

$10$ vibrations of longer pendulum= $11$ vibrations of shorter pendulum

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

Assertion (A): A wooden cube of side a floats in a non viscous liquid of density r. When it is slightly pressed and released, then it executes SHM
Reason (R): The net force responsible for SHM is the resultant of buoyancy force and true weight of the body.

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true and R is not the correct explanation of A

  3. A is true and R is false

  4. A is false and R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a wooden cube of side a floats in a non viscous liquid of density r and if it is slightly pressed and released it executes SHM because of the buoyancy force which is acting on the cube in upward direction. And hence the answer. 

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

A body is broken into two parts of masses $m _1$ and $m _2$ These parts are then separated by a distance r ,What is the value of $m _1/m _2$ so that the gravitational force has maximum possible value?

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

Let the mass of the body is $m$

$\therefore \,{m _1} + {m _2} = m$
${F _G} = \frac{{G{m _1}{m _2}}}{{{r^2}}}$
${F _G} = \frac{{G{m _1}\left( {m - {m _1}} \right)}}{{{r^2}}}$
for ${F _G} \to \max \,\frac{{d\left( {{F _G}} \right)}}{{d{m _1}}} = 0$
$ = \frac{G}{{{r^2}}}\left( {m - 2{m _1}} \right) = 0$
$ = {m _1} = \frac{m}{2}$
${m _2} = \frac{m}{2}$
$\therefore {m _1}/{m _2} = 1:1$
Hence$,$ optin $(A)$ is correct$.$ 

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

A person normally weighing 60kg stands on a platform which oscillates up and down simple harmonically with a frequency $2Hz$ and an amplitude $5cm$.if a machine on the platform gives the person's weight,then consider the following statements :

  1. The maximum reading of machine will be $108$kg
  2. The maximum reading of machine will be $90kg$
  3. The minimum reading of machine will be $12kg$

  4. The minimum reading of the machine will be zero correct statements are:
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Maximum a=$w^2A$
$=(4\pi)^2\times 0.05$
$=16 \pi^2\times 0.05$
$=8$
$mg=60 \Rightarrow m=\dfrac{60}{10}=6$
$W _{max}=m(g+a)$
$=6\times (18)$
$=108kg$
$W _{min}=6(g-a)
$=12kg

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

If a tunnel is cut at any orientation through earth, then in what time will, a ball released from one end, reach the other end (neglect the rotation of the earth) ?

  1. 84.6 minutes

  2. 42.3 minutes

  3. 8 minutes

  4. depends on orientation
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Force and to rque 
As we know
g depth = $g\left ( 1-\frac{d}{R} \right )$
$g _{d} = g\left ( \frac{R-d}{R} \right )\because (R-d)=x$
$g _{d}=\frac{g}{R}x$
$F _{r}=-mg _{d}$
$F _{R}=\frac{-mgx}{R}\Rightarrow F = -KX$
$T = 2\pi \sqrt{\frac{m}{k}}\rightarrow $ As we know
$K = \frac{mg}{R}$
$T= 2\pi \sqrt{\frac{mR}{mg}}$
$T= 2\pi \sqrt{\frac{R}{g}}$
$T = 84.6 ,minutes $
$A\rightarrow B;t=\frac{T}{2}=\frac{84.6}{2}=43.2$
Multiple choice physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

A ball is in simple harmonic motion in a tunnel through center of the earth. Magnitude of gravitational force acting on the ball of radius $ y _o $ ,when it is at a distance $x$ from mean position is :

  1. $\dfrac{GMm}{R^{3}}x$
  2. $\dfrac{GMm}{\left [ (R-y _{0})^{2})+x^{2} \right ]}$
  3. $\dfrac{GMm}{R^{3}}\left [ (R-y _{0})^{2}+x^{2} \right ]^{1/2}$
  4. $\dfrac{GMmR^{2}}{\left [ (R-y _{0})^{2}+x^{2} \right ]^{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a tunnel through the Earth, the gravitational force inside a uniform sphere is proportional to the distance from the center. The effective distance from the center for a point at distance x from the mean position (center) is r = sqrt((R-y_0)^2 + x^2). The force is F = (GMm/R^3) * r.

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

A solid cube of side $a$ and density $\rho _{0}$ floats on the surface of a liquid of density $\rho $. If the cube is slightly pushed downward, then it oscillates simple harmonically with a period of:

  1. $\displaystyle 2\pi \sqrt{\frac{\rho _{0}}{\rho }\frac{a}{g}}$
  2. $\displaystyle 2\pi \sqrt{\frac{\rho }{\rho _{0}}\frac{a}{g}}$
  3. $\displaystyle 2\pi \sqrt{\frac{a}{\left ( 1-\frac{\rho }{\rho _{0}} \right )g}}$
  4. $\displaystyle 2\pi \sqrt{\frac{a}{\left ( 1+\frac{\rho }{\rho _{0}} \right )g}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle T=2\pi \sqrt{\frac{m}{k}}$
Here Mass of the cube is $m=\left ( \rho _{0} \right )\left ( a^{3} \right )$
Since the cube is pushed very slightly so the part dipped into the water is very negligible 

Spring constant for a negligible distance  $k=\rho _{1} \times $  bottom surface area of cube $\times g=\rho a^{2}g$
$\therefore $   $\displaystyle T=2\pi \sqrt{\frac{\rho _{0}a}{\rho g}}$

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

A ball is in simple harmonic motion in a tunnel through center of the earth. Total force that acts on the ball when it is at a distance $x$ from mean position is :

  1. $\dfrac{GMm}{R^{2}}$
  2. $\dfrac{GMm}{R^{3}}x$
  3. $\dfrac{GMm}{(R-y _{0})^{2}+x^{2}}$
  4. $\dfrac{GMmR^{2}}{\left [ (R-y _{0})^{2}+x^{2} \right ]^{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


Let $M$ be the total mass of the earth. At any position of $x$ let the mass be $M'$.

Mass = density x volume

So,

$\dfrac { { M }^{ ' } }{ M } =\dfrac { \rho \times \dfrac { 4 }{ 3 } \pi { x }^{ 3 } }{ \rho \times \dfrac { 4 }{ 3 } \pi { R }^{ 3 } } =\dfrac { { x }^{ 3 } }{ { R }^{ 3 } } \\ \Rightarrow { M }^{ ' }=\dfrac { { x }^{ 3 } }{ { R }^{ 3 } } M$


So the force on the ball is given by 


${ F } _{ x }=\dfrac { G{ M }^{ ' }m }{ { x }^{ 2 } } =\dfrac { Gm }{ { x }^{ 2 } } \left( \dfrac { { x }^{ 3 } }{ { R }^{ 3 } } M \right) =\dfrac { G{ M }m }{ { R }^{ 3 } } x$

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

Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance $h$ directly above the tunnel, the motion of the particle is

  1. Simple harmonic

  2. Parabolic

  3. Oscillatory

  4. Periodic
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

When a particle is dropped from a height $h$ above the centre of tunnel.
$(i)$ It will oscillate, through the earth to a height $h$ on both sides
$(ii)$ The motion of particle is periodic
$(iii)$ The motion of particle will not be $SHM$.