Simple Harmonic Motion
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 \mathrm{~kg}$, the angular frequency is $\omega_{1}$. When the mass block is $2 \mathrm{~kg}$ the angular frequency is $\omega_{2}$. The ratio $\omega_{2} / \omega_{1}$ is
A particle executes simple harmonic motion between $x=-A$ and $x=+A$. If time taken by particle to go from $x=0$ to $\frac{A}{2}$ is 2 s; then time taken by particle in going from $x=\frac{A}{2}$ to A is
T is the time period of simple pendulum on the earth's surface. Its time period becomes $x$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be :
At a given point of time the value of displacement of a simple harmonic oscillator is given as $\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)$. If amplitude is $40 \mathrm{~cm}$ and kinetic energy at that time is $200 \mathrm{~J}$, the value of force constant is $1.0 \times 10^{x} ~\mathrm{Nm}^{-1}$. The value of $x$ is ____________.
Explanation:
$x = A \sin(\omega t + \phi)$
At the given time, we have:
$\omega t + \phi = 30^\circ$
Given the amplitude $A = 40 \,\text{cm}$ and the displacement $x = 40 \times \frac{\sqrt{3}}{2} \,\text{cm} = 20\sqrt{3} \,\text{cm}$, we can write the kinetic energy, $KE$, as:
$KE = \frac{1}{2}k(A^2 - x^2) = 200$
Now, substitute the values for $A$ and $x$:
$200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 \times 100}\right)$
Simplify the equation:
$400 \times 100 \times 100 = k \times 400$
Solve for the force constant, $k$:
$k = 10^4 \,\text{Nm}^{-1}$
Given that the force constant is expressed as $k = 1.0 \times 10^x \,\text{Nm}^{-1}$, comparing the values, we get:
$1.0 \times 10^x = 10^4$
Thus, the value of $x$ is $4$.
A rectangular block of mass $5 \mathrm{~kg}$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum force exerted by spring on block is _________ N
Explanation:
To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.
First, let's find the angular frequency $\omega$:
$\omega = \frac{2\pi}{T}$
where $T = 3.14\,\mathrm{s}$ is the time period.
$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$
Now, let's find the maximum velocity $v_{max}$ of the block:
$v_{max} = \omega A$
where $A = 1\,\mathrm{m}$ is the amplitude.
$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$
Next, we can find the spring constant $k$ using the mass of the block $m = 5\,\mathrm{kg}$ and the angular frequency $\omega$:
$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$
$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$
Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:
$F_{max} = kA$
$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$
The maximum force exerted by the spring on the block is $20\,\mathrm{N}$.
A simple pendulum with length $100 \mathrm{~cm}$ and bob of mass $250 \mathrm{~g}$ is executing S.H.M. of amplitude $10 \mathrm{~cm}$. The maximum tension in the string is found to be $\frac{x}{40} \mathrm{~N}$. The value of $x$ is ___________.
Explanation:
Given the amplitude $A$ and the length $l$ of the pendulum, we can find the maximum angular displacement $\theta_0$:
$ \sin \theta_0 = \frac{A}{l} = \frac{10}{100} = \frac{1}{10} $
By conservation of energy, the following equation holds:
$ \frac{1}{2} m v^2 = m g l(1 - \cos \theta) $
The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:
$ \begin{aligned} & T - mg = \frac{m v^2}{l} \ & \Rightarrow T = mg + \frac{m v^2}{l} \end{aligned} $
Substituting the conservation of energy equation, we get:
$ \begin{aligned} & T = mg + 2 m g(1 - \cos \theta) \\\\ & = mg\left[1 + 2\left(1 - \sqrt{1 - \sin^2 \theta}\right)\right] \\\\ & = mg\left[3 - 2 \sqrt{1 - \frac{1}{100}}\right] \\\\ & = \frac{250}{1000} \times 10\left[3 - 2\left(1 - \frac{1}{200}\right)\right] = \frac{101}{40} \\\\ & \therefore x = 101 \end{aligned} $
So, the value of $x$ is 101.
The amplitude of a particle executing SHM is $3 \mathrm{~cm}$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is: __________ $\mathrm{cm}$
Explanation:
$ \begin{aligned} & K=1.25 U \\\\ & \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\ & \Rightarrow \frac{9}{5} K=K_{\max } \\\\ & \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\ & \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} \\\\ & \Rightarrow 9\left(A^{2}-x^{2}\right)=5 A^{2} \\\\ & \Rightarrow x^{2}=\frac{4 A^{2}}{9} \\\\ & \Rightarrow x=\frac{2 A}{3} \\\\ & \Rightarrow x=2 \mathrm{~cm} \end{aligned} $
In the figure given below, a block of mass $M=490 \mathrm{~g}$ placed on a frictionless table is connected with two springs having same spring constant $\left(\mathrm{K}=2 \mathrm{~N} \mathrm{~m}^{-1}\right)$. If the block is horizontally displaced through '$\mathrm{X}$' $\mathrm{m}$ then the number of complete oscillations it will make in $14 \pi$ seconds will be _____________.
Explanation:
$=2 \times 2=4 \mathrm{~N} / \mathrm{m}$
$ \mathrm{m}=490 \mathrm{gm} $
$ \begin{aligned} & =0.49 \mathrm{~kg} \end{aligned} $
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}$
$ =2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10} $
No. of oscillation in the $14 \pi$ is
$ \mathrm{N}=\frac{\text { time }}{\mathrm{T}}=\frac{14 \pi}{7 \pi / 10}=20 $
Explanation:
$4{v^2} = 50 - {x^2}$
or $v = {1 \over 2}\sqrt {50 - {x^2}} $
Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}} $
$ \Rightarrow \omega = {1 \over 2}$
& $A = \sqrt {50} $
so ${{2\pi } \over T} = {1 \over 2}$
$ \Rightarrow T = 4\pi \sec $
$ = 4 \times {{22} \over 7}\sec $
$T = {{88} \over 7}\sec $
so $x = 88$
The general displacement of a simple harmonic oscillator is $x = A\sin \omega t$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t = {T \over \beta }$. The value of $\beta$ is ______________.
Explanation:
$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$
So, ${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$
This value will be maximum when $\sin 2\omega t = 1$
or $2\omega t = {\pi \over 2}$
$2 \times {{2\pi } \over T}t = {\pi \over 2}$
$ \Rightarrow t = {T \over 8}$
So $\beta = 8$
A particle of mass 250 g executes a simple harmonic motion under a periodic force $\mathrm{F}=(-25~x)\mathrm{N}$. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.
Explanation:
$F = - 25x$
$.250{{{d^2}x} \over {d{t^2}}} = - 25x$
${{{d^2}x} \over {d{t^2}}} = - 100x$
$ \Rightarrow \omega = 10$ rad/sec
& $\omega A = {v_{\max }}$
$10\,A = 4$
$ \Rightarrow A = 0.4$ m
$ = 40$ cm
A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3 kg the period of the oscillation increases by one second, the value of mass m is ___________ kg.
Explanation:
Finally
$ 2 \pi \sqrt{\frac{m+3}{k}}=1+1=2 $
Equation $\frac{(1)}{(2)}$ gives
$\sqrt{\frac{m}{m+3}}=\frac{1}{2}$
$\therefore m=1 \mathrm{~kg}$
A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt x}$ in SI unit. The value of $x$ is ____________.
Explanation:
So $T=2 \pi \sqrt{\frac{m}{K_{n e t}}}$
$ \begin{aligned} & =2 \pi \sqrt{\frac{2}{40}} \\\\ & =\frac{\pi}{\sqrt{5}} \end{aligned} $
$\therefore x=5$
The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by :
Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -
(consider radius of earth $R_{E}=6400 \mathrm{~km}$ and $\mathrm{g}$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )
When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :
In figure $(\mathrm{A})$, mass '$2 \mathrm{~m}^{\text {' }}$ is fixed on mass '$\mathrm{m}$ ' which is attached to two springs of spring constant $\mathrm{k}$. In figure (B), mass '$\mathrm{m}$' is attached to two springs of spring constant '$\mathrm{k}$' and '$2 \mathrm{k}^{\prime}$. If mass '$\mathrm{m}$' in (A) and in (B) are displaced by distance '$x^{\prime}$ horizontally and then released, then time period $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ corresponding to $(\mathrm{A})$ and (B) respectively follow the relation.
The motion of a simple pendulum executing S.H.M. is represented by the following equation.
$y = A\sin (\pi t + \phi )$, where time is measured in second. The length of pendulum is
Motion of a particle in x-y plane is described by a set of following equations $x = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m$ and $y = 4\sin (\omega t)\,m$. The path of the particle will be :
The equation of a particle executing simple harmonic motion is given by $x = \sin \pi \left( {t + {1 \over 3}} \right)m$. At t = 1s, the speed of particle will be
(Given : $\pi$ = 3.14)
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :
Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with ${g \over 6}$ vertically upwards then the time period will be :
(Where g = acceleration due to gravity)
Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :
The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10 \mathrm{~s}$. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{x}$ s. The value of $x$ will be ________.
Explanation:
$\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{F}_{\mathrm{B}}$
$\mathrm{g}^{\prime}=\frac{\mathrm{mg}-\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}$
$=\frac{\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{w}} \mathrm{Vg}}{\rho_{\mathrm{B}} \mathrm{V}}$
$=\left(\frac{\rho_{\mathrm{B}}-\rho_{\mathrm{w}}}{\rho_{\mathrm{B}}}\right) \mathrm{g}$
$=\frac{5-1}{5} \times \mathrm{g}$
$=\frac{4}{5} \mathrm{~g}$
We know, $T =2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
$\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}^{\prime}}}=\sqrt{\frac{\mathrm{g}}{5} \mathrm{~g}}=\sqrt{\frac{5}{4}}$
$\mathrm{~T}^{\prime}=\mathrm{T} \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}$
$\mathrm{~T}^{\prime}=5 \sqrt{5}$
The potential energy of a particle of mass $4 \mathrm{~kg}$ in motion along the x-axis is given by $\mathrm{U}=4(1-\cos 4 x)$ J. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) s$. The value of $\mathrm{K}$ is _________.
Explanation:
$U = 4(1 - \cos 4x)$
$ \Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$
$ = - 16\sin 4x$
as small x
$F = - 16(4x) = - 64x \equiv - kx$
$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$
$ \Rightarrow K = 2$
A mass $0.9 \mathrm{~kg}$, attached to a horizontal spring, executes SHM with an amplitude $\mathrm{A}_{1}$. When this mass passes through its mean position, then a smaller mass of $124 \mathrm{~g}$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$, then the value of $\alpha$ will be ___________.
Explanation:
$(0.9){A_1}\sqrt {{K \over {0.9}}} = (0.9 + 0.124){A_2}\sqrt {{K \over {0.9 + 0.124}}} $
${{{A_1}} \over {{A_2}}} = \sqrt {{{0.9 + 0.124} \over {0.9}}} $
$ = \sqrt {{{1.024} \over {0.9}}} $
$ = {\alpha \over {\alpha - 1}}$
$\alpha = 16$
As per given figures, two springs of spring constants $k$ and $2 k$ are connected to mass $m$. If the period of oscillation in figure (a) is $3 \mathrm{s}$, then the period of oscillation in figure (b) will be $\sqrt{x}~ s$. The value of $x$ is ___________.
Explanation:
For case (a),
${K_{eq}} = {{2K} \over 3}$
For case (b),
${K_{eq}} = 3K$
$\because$ $T = 2\pi \sqrt {{m \over K}} $
$\therefore$ ${{{T_a}} \over {{T_b}}} = \sqrt {{{{K_b}} \over {{K_a}}}} $
${3 \over {{T_b}}} = \sqrt {{{3K \times 3} \over {2K}}} = {3 \over {\sqrt 2 }}$
${T_b} = \sqrt 2 $
$x = 2$
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt{x}$ cm. The value of x is _____________.
Explanation:
$v = \omega \sqrt {{A^2} - {y^2}} $
$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $
$ \Rightarrow 9 \times 75 = {(A')^2} - 25$
$ \Rightarrow A' = \sqrt {28 \times 25} $ cm
$ \Rightarrow x = 700$
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60$^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms$-$1. (if g = 10 m/s2)
Explanation:
${1 \over 2}m{v^2} = mgl(1 - \cos \theta )$
$ \Rightarrow v = \sqrt {2gl(1 - \cos \theta )} $
$ = \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}} $
$ = 5$ m/s
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.
Explanation:
A = 8 cm
T = 6 s
$A\cos \left( {{{2\pi t} \over T}} \right) = {A \over 2}$
$ \Rightarrow {{2\pi t} \over T} = {\pi \over 3}$
or $t = {T \over 6} = 1\,s$
(1) Potential energy is always equal to its K.E.
(2) Average potential and kinetic energy over any given time interval are always equal.
(3) Sum of the kinetic and potential energy at any point of time is constant.
(4) Average K.E. in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below :
The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :
Statement I : A second's pendulum has a time period of 1 second.
Statement II : It takes precisely one second to move between the two extreme positions.
In the light of the above statements, choose the correct answer from the options given below :
