Simple harmonic motion - class-IX

In simple harmonic motion, the average kinetic energy and the average potential energy over one complete period are equal, both being half of the total energy.

At the extreme position, kinetic energy is zero and we know to throw out S.H.M.

Let the equation of $S.H.M$ is:-

Given,

Given,

The equation is d^2x/dt^2 + 16x = 0. Comparing this to d^2x/dt^2 + w^2x = 0, we get w^2 = 16, so w = 4 rad/s. Given mass m = 2g = 0.002 kg, the force constant K = m * w^2 = 0.002 * 16 = 0.032 N/m.

$\begin{array}{l} m=3bg,, A=13,T=125 \ displacementx\left( t \right) =13\sin  \left( { \frac { { 2\pi t } }{ T }  } \right)  \end{array}$

$\begin{array}{l} \sin  \omega t-\cos  \omega t \ =\sqrt { 2 } \left[ { \frac { 1 }{ { \sqrt { 2 }  } } \sin  \omega t-\frac { 1 }{ { \sqrt { 2 }  } } \cos  \omega t } \right]  \ =\sqrt { 2 } \left[ { \sin  \omega t\times \cos  \frac { \pi  }{ 4 } -\cos  \omega t\times \sin  \frac { \pi  }{ 4 }  } \right]  \ =\sqrt { 2 } \sin  \left( { \omega t-\frac { \pi  }{ 4 }  } \right)  \ this, , function, , represents, , SHM, , as, , it, , can, , be, , written, , in, , the, , form: \ a\sin  \left( { \omega t+\phi  } \right)  \ its, , period, , is,, , \frac { { 2\pi  } }{ \omega  }  \end{array}$

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

We know that for SHM, $x=A\sin(\omega t + \theta)$ and $F=kx=kA\sin(\omega t + \theta)$
Thus it's a sine curve.

$a=\omega^{2}x$
So, $a _{max}=$ $\omega^{2}A$
We know that $\omega=2\pi  f$
So, $a _{max}=\dfrac{(2\pi\times 2)^{2}\times 5}{100}$
Case I:
$N-mg=ma _{max}$
$N=m(a _{max}+g)$
$=60(10+\dfrac{16\times \pi^{2}\times 5}{100})$
$=1080 $ 

$ N=108$ kg-wt

Case II:
$mg-N=ma _{max}$
or, $N=mg-ma _{max}$
$=60(10-8)$
$=120\ N=12$ kg-wt

$T=\dfrac{ \pi}{2}$
$V _{max}=1 m/sec$
$\omega =\dfrac{2\pi}{T}$
$V=4  rad/sec$
$V _{max=}A\omega$
$A\times 4=1$
$A=\dfrac {1}{4}$
$a=\omega^{2}x$
$F= m\omega^{2}x$
$x=A   cos   60^o$
$\therefore x=\dfrac {A}{2}$
$F=0.5\times (4)^{2}\times \dfrac {1}{4}\times \dfrac{1}{2}$
$F=1N$

In SHM a constant force brings no effective change in the motion but the mean position accelerates.

Guitar string is a standing wave as both the ends of guitar string are fixed.

$tan \phi = \dfrac{\omega L- \dfrac{1}{\omega C}}{R}$

If $\omega < \omega _0 \Longrightarrow$ circuit is capacitive

$\Longrightarrow$ it leads voltage
$\Longrightarrow$ velocity leads force.
if $\omega = \omega _0 \phi = 0$
if $\omega > {\omega} _0$ velocity lags behind force.

The fundamental frequency of a stretched string is given by f = (1/2L) * sqrt(T/mu). Given L = 1 m, mass m = 5 x 10^-4 kg so linear mass density mu = 5 x 10^-4 kg/m, and tension T = 20 N. Plugging these in yields f = (1/2) * sqrt(20 / (5 x 10^-4)) = (1/2) * sqrt(40000) = (1/2) * 200 = 100 Hz.

$V^2=108-9x^2$

The equation is d^2x/dt^2 + pi^2 * x = 0. Comparing to d^2x/dt^2 + w^2 * x = 0, we get w^2 = pi^2, so w = pi. Since w = 2 * pi * f, then pi = 2 * pi * f, which gives f = 1/2 Hz.

Total force on the particle will be
$F=mg+m\dfrac { d^{ 2 }y }{ dt^{ 2 } } $
since $y=a(1-\cos { \omega t } )\\ \Rightarrow \dfrac { dy }{ dt } =a\omega \sin { \omega t } \\ \Rightarrow \dfrac { d^{ 2 }y }{ dt^{ 2 } } =a\omega ^{ 2 }\cos { \omega t } $
$\Rightarrow F=mg+ma\omega ^{ 2 }\cos { \omega t } $

The frequency of seconds pendulum is $= \dfrac{1}{2} = 0.5$ HZ

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. 

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,

The time period of simple pendulum id given by

$T _0=2\pi\sqrt{\cfrac{l}{g}}\T^1=2\pi\sqrt{\cfrac{l+l\times2/100}{g}}\ \cfrac{T _0}{T^1}=\cfrac{\sqrt{100}}{\sqrt{102}}\ T^1=\cfrac{\sqrt{102}}{\sqrt{100}}T _0\T^1=1.0099T _0\approx  1.01T _0\Loss=(1.01-1)T _0=0.01T _0$

T = 2*pi * sqrt(L/g). dT/T = 0.5 * dL/L. Given dL/L = 0.02, dT/T = 0.5 * 0.02 = 0.01. Time lost per day = (dT/T) * 86400s = 0.01 * 86400 = 864 seconds.

The differential equation for simple harmonic motion is d^2x/dt^2 + omega^2 x = 0. For a seconds pendulum, the time period T = 2 seconds, so angular frequency omega = 2pi/T = 2pi/2 = pi. Substituting omega^2 into the equation gives d^2x/dt^2 + pi^2 x = 0.

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) $
  $=2\sqrt { 2 } s$

Time period of pendulum is:

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.

We know that the time period of a second pendulum is $2s$ .

The time period of a simple pendulum is
$T = 2 \pi \sqrt{\dfrac{L}{g}}$
where L is the length of the pendulum.
or $ L = \dfrac{gT^2}{4 \pi^2}$
The time period of the simple pendulum which ticks seconds is $2$s.
$\therefore T = 2s$
Substituting in (i), we get

Correct answer= B

When a rigid body is suspended vertically, and it oscillates with a small amplitude under the action of the force of gravity, the body is known as compound pendulum. Thus the periodic time of a compound pendulum is minimum when the distance between the point of suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.

The correct option is (c)

The amplitude decay of a pendulum in a viscous medium follows the equation A = A_0 * exp(-bt/2m). The damping constant b is proportional to the viscosity eta. Since the ratio of viscosities is 1.3, the decay constant in CO2 is 1.3 times that in air. By comparing the time taken to reach half amplitude, the result is calculated as 161 seconds.

Time taken for 1 oscillation is 2 s
Time taken for 20 oscillations is $(2 \times 20) = 40  s$

$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
First distance of comfrom suspension point will increase then decrease.

Seconds pendulum takes 2 $sec$ to make one complete oscillation. 

A seconds pendulum is a pendulum whose period is precisely two seconds

The fractional change in time period is given by delta T / T = 1/2 * alpha * delta theta. With alpha = 1210^-6 and delta theta = 20 degrees, the fractional change is 1.210^-4. Multiplying by the number of seconds in a day (86400), we get approximately 10.36 seconds.

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

$T, =, 2 \pi,\sqrt{\displaystyle \frac{L}{g}}, \Rightarrow, T^2, =, 4 \pi^2, \times, \displaystyle \frac{L}{g}$ .. (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 length $121cm$ and $100cm$.

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. 

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

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.

$\displaystyle T=2\pi \sqrt{\frac{m}{k}}$
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 

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 

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

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 $SHM$.

$F = mg\left ( \left ( 1- \dfrac{a}{2} \right )- \dfrac{\beta h}{2}\right )$

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}}$
$\sqrt{\frac{900}{650\times .54}}=\sqrt{\frac{g}{l}}$
$l=g\frac{650\times .54}{900}$
$l=.06\times 65$
$=39 cm$