Tag: example of simple harmonic motion

Questions Related to example of simple harmonic motion

Find the length of a simple pendulum such that its time period is $2\ s$.

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

  1. $99.4\ cm$

  2. $89.4\ cm$

  3. $79.4\ cm$

  4. $109.4\ cm$

Show answer
Correct Option: A
Explanation:

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


$T^2\, =\, 4\pi ^2\, \displaystyle \frac {L}{g}$
$\Rightarrow\, 2^2\, =\, 4\, \times\, 3.14\, \times\, 3.14\, \times\, \displaystyle \frac {L}{9.8}$
$\Rightarrow\, L\, =\, \displaystyle \frac {4\, \times\, 9.8}{4\, \times\, 3.14\, \times\, 3.14}\, m\, =\, 0.994\, m\, =\, 99.4\, cm$

A desktop toy pendulum swings back and forth once every $1.0 s$. How long is this pendulum?

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

  1. $0.25\, m$

  2. $0.50\, m$

  3. $0.15\, m$

  4. $0.30\, m$

Show answer
Correct Option: A
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.)

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

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

  1. $0.25\ m$

  2. $0.50\ m$

  3. $0.25\ cm$

  4. $0.25\ mm$

Show answer
Correct Option: A
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.)

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? 

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

  1. 10 vibrations

  2. 11 vibrations

  3. 21 vibrations

  4. 20 vibrations

Show answer
Correct Option: A
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

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.

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

  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

Show answer
Correct Option: A
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. 

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?

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

  1. $1 : 1$

  2. $1 : 2$

  3. $2: 1$

  4. $4 : 3$

Show answer
Correct Option: A
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$.$ 

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 :

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

  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:

Show answer
Correct Option: A,C
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

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) ?

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

  1. 84.6 minutes

  2. 42.3 minutes

  3. 8 minutes

  4. depends on orientation

Show answer
Correct Option: B
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 $
$A\rightarrow B;t=\frac{T}{2}=\frac{84.6}{2}=43.2$

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 :

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

  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. $\dfrac{GMmR^{2}}{\left [ (R-y _{0})^{2}+x^{2} \right ]^{2}}$

Show answer
Correct Option: C

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:

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

  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. $\displaystyle 2\pi \sqrt{\frac{a}{\left ( 1+\frac{\rho }{\rho _{0}} \right )g}}$

Show answer
Correct Option: A
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 $   $\displaystyle T=2\pi \sqrt{\frac{\rho _{0}a}{\rho g}}$