Tag: a few applications of linear shm

Questions Related to a few applications of linear shm

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:

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

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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 $   $\displaystyle T=2\pi \sqrt{\frac{\rho _{0}a}{\rho g}}$

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 :

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

  1. $\dfrac{GMm}{R^{2}}$

  2. $\dfrac{GMm}{R^{3}}x$

  3. $\dfrac{GMm}{(R-y _{0})^{2}+x^{2}}$

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

Show answer
Correct Option: B
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 


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

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

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

  1. Simple harmonic

  2. Parabolic

  3. Oscillatory

  4. Periodic

Show answer
Correct Option: C,D
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 $SHM$.

A small ball of density $\rho _{0}$ is released from rest from the surface of a liquid whose density varies with depth $h$ as $\rho =\dfrac{\rho _{0}}{2}(a+\beta h)$Mass of the ball is $m$.Select the most appropriate option:

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

  1. The particle will execute SHM.

  2. The maximum speed of the ball is $\dfrac{2-a}{\sqrt{2\beta }} $.

  3. Both (a) and (b) are correct.

  4. Both (a) and (b) are wrong.

Show answer
Correct Option: A
Explanation:
$\rho = \dfrac {\rho _{o}}{2} (a + \beta h)$
$V = \dfrac{m}{\rho _{o}}$
Downward force, $F = mg - \dfrac{m}{\rho _{o}}\dfrac{\rho _{o}}{2}(a+\beta h)g$
$F = mg\left ( \left ( 1- \dfrac{a}{2} \right )- \dfrac{\beta h}{2}\right )$
So F is proportional to displacement from center position which is $mg (1 - \dfrac{a}{2})$ below surface
Therefore, the motion is SHM.
A = $mg (1 - \dfrac{a}{2})$
$k = \dfrac{mg \beta}{2}$
Also, $k = m\omega^{2}$
=> $\omega = \sqrt{\dfrac{g\beta}{2}}$
$v _{max} = \omega A = \dfrac{(2-a) \sqrt{g}}{\sqrt{2} \beta}$
Answer A

A cylindrical block of wood $(density=650 kg m^{-3})$, of base area $30 cm^2$ and height $54 cm$, floats in a liquid of density $900 kg$ $m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly)

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

  1. 52 cm

  2. 26 cm

  3. 39 cm

  4. 65 cm

Show answer
Correct Option: C
Explanation:

As block is floating it's weight should be equal to buoyancy force 
$\rho _{wood} V _{cylinder} g=\rho _{liquid} V _{displaced} g$
$V _{displaced} =\frac{\rho _{wood} V _{cylinder} }{\rho _{liquid}} $
$V _{displaced} =\frac{\rho _{wood} V _{cylinder} }{\rho _{liquid}} $...(i)
After displacing by small distance x, the net force on cylinder will be
$F=Buoyancy-w=\rho _{liquid}( V _{displaced} +A _{cylinder}\Delta x) g- \rho _{wood} V _{cylinder} g$
$F=\rho _{liquid}  V _{displaced} g+\rho _{liquid} A _{cylinder}\Delta x g-\rho _{wood} V _{wood} g$ the net force on cylinder becomes 
from equation (i) $\rho _{wood} V _{cylinder} g=\rho _{liquid} V _{displaced} g$
$F=\rho _{liquid} A _{cylinder}\Delta x g$
$ma=\rho _{liquid} A _{cylinder}\Delta x g$
$a=\frac{\rho _{liquid} A _{cylinder}}{m}\Delta x $
$a = \omega^2 \Delta x $
$\omega^2 =\frac{\rho _{liquid} A _{cylinder}}{\rho _{cylinder} V _{cylinder}}$
$\omega^2 =\frac{\rho _{liquid} A _{cylinder}}{\rho _{wood} A _{cylinder} h _{cylinder}}$
$\omega^2 =\frac{\rho _{liquid}}{\rho _{wood} h _{cylinder}}$
$\omega^2 =\frac{900}{650\times .54 }$
This should be equal to angular frequency of simple pendulum
$ \omega=\sqrt{\frac{g}{l}}$
$\sqrt{\frac{900}{650\times .54}}=\sqrt{\frac{g}{l}}$
$l=g\frac{650\times .54}{900}$
$l=.06\times 65$
$=39 cm$