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Physics
Oscillations
63 questions, each with its answer.
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For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by:$K.E. = \frac{1}{2}m\omega^2 A^2 \cos^2\omega t$ (oscillation starts from the extreme position) K.E. is always non-negative and varies as $\cos^2\omega t$, so it repeats twice in one time period (period of K.E. = $\frac{T}{2}$).
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Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^2 = 9.8$ and $g = 9.8\ m/s^2$)Time period $T = \frac{60}{30} = 2$ s $T = 2\pi\sqrt{\frac{l}{g}}$ $2 = 2\pi\sqrt{\frac{l}{\pi^2}}$ (since $g = \pi^2 = 9.8$) $l \approx 1$ m
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The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob = 20 g)Total energy $= \frac{1}{2}KA^2 = \frac{1}{2}m\omega^2A^2 = 0.02$ J At equilibrium, all energy is kinetic: $\frac{1}{2}mv_{max}^2 = 0.02$ J $v_{max}^2 = \frac{0.04}{20 \times 10^{-3}} = 2$ $v_{max} = 1.41$ m/s
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A cylindrical cork of uniform density floats in a liquid of density $\rho_1$. If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density $\rho_2$, then the similar oscillation has time period 2T. The value of $\rho_2/\rho_1$ is:For a small depression x, the restoring force is $F=-\rho gAx$. Comparing with $F=-kx$ gives $k=\rho Ag$. $T=2\pi\sqrt{\frac{m}{\rho Ag}} \Rightarrow T\propto\frac{1}{\sqrt{\rho}}$. $\frac{T}{2T}=\sqrt{\frac{\rho_2}{\rho_1}} \Rightarrow \frac{\rho_2}{\rho_1}=\frac{1}{4}$.
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Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k Nm$^{-1}$. At a given instant, the extension of the spring is x meter and the speed of the particle is v ms$^{-1}$. On the x - v plane, if the graph of v as a function of x is a circle, then the correct option is:Total energy $E=\frac{1}{2}mv^2+\frac{1}{2}kx^2 \Rightarrow \frac{v^2}{2E/m}+\frac{x^2}{2E/k}=1$. Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the curve is a circle when $a^2=b^2 \Rightarrow \frac{2E}{m}=\frac{2E}{k} \Rightarrow k=m$.
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In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
Since mass $m(t)$ decreases, $\omega = \sqrt{k/m}$ increases, while lost sand carries away mechanical energy, steadily decreasing amplitude $A(t)$.
For a spring-mass system, the angular frequency is $\omega(t) = \sqrt{k/m(t)}$. As sand leaks out, total mass $m(t)$ decreases, which implies that $\omega(t)$ progressively increases. Simultaneously, as sand drops vertically out of the moving box, it carries away kinetic energy without an opposing reaction force, so the total mechanical energy of the remaining oscillator decreases, causing the average amplitude $A(t)$ to steadily decrease. Thus, graph (2) correctly shows $\omega(t)$ rising and $A(t)$ declining. -
Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$ and $k_2$ respectively, oscillate vertically. If their maximum speeds are the same, the ratio $(A_Q / A_P)$ of the amplitude $A_Q$ of mass Q to the amplitude $A_P$ of mass P is:
$v_{max} = \omega A = \sqrt{k/m} A$. Equating maximum speeds gives $\sqrt{k_1} A_P = \sqrt{k_2} A_Q \implies A_Q / A_P = \sqrt{k_1 / k_2}$.
For simple harmonic oscillation of a mass-spring system, the maximum velocity is $v_{max} = \omega A = \sqrt{\frac{k}{m}} A$. Given that both masses are identical ($m_P = m_Q = m$) and have equal maximum velocities: $v_{max,P} = v_{max,Q} \implies \sqrt{\frac{k_1}{m}} A_P = \sqrt{\frac{k_2}{m}} A_Q$. Rearranging for the ratio of amplitudes: $\frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$. -
If $x=5\sin\left(\pi t + \frac{\pi}{3}\right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are;Comparing with $x = A\sin(\omega t + \phi)$: A = 5 m and $\omega = \pi$ rad $s^{-1}$ $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2$ s So amplitude = 5 m and time period = 2 s.
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If the mass of the bob in simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:Time period of a pendulum $T = 2\pi\sqrt{\frac{l}{g}}$ (independent of the mass of the bob) $T' = 2\pi\sqrt{\frac{l}{2g}} = \frac{1}{\sqrt{2}}T$ $\frac{x}{2}T = \frac{1}{\sqrt{2}}T \Rightarrow \sqrt{2}x = 2$ $x = \frac{2}{\sqrt{2}} = \sqrt{2}$
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A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement$U=\dfrac{1}{2}kx^2$, $K=\dfrac{1}{2}kA^2-\dfrac{1}{2}kx^2$. Setting $U=K$: $\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2-\dfrac{1}{2}kx^2\Rightarrow x^2=\dfrac{A^2}{2}\Rightarrow x=\dfrac{A}{\sqrt2}$.
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The two-dimensional motion of a particle, described by $\vec{r}=(\hat{i}+2\hat{j})A\cos\omega t$ is a/an: A. parabolic path B. elliptical path C. periodic motion D. simple harmonic motion. Choose the correct answer from the options given below:Since $x=A\cos\omega t$ and $y=2A\cos\omega t$, we get $y=2x$, a straight line — not parabolic or elliptical. $\dfrac{d\vec r}{dt}=-(\hat i+2\hat j)\omega A\sin\omega t$ and $\dfrac{d^2\vec r}{dt^2}=-\omega^2\vec r$, i.e. $\vec a=-\omega^2\vec r$, which is the defining condition of SHM, and SHM is periodic. So only C and D apply.
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The x–t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=2\ s$ is:
From the graph $A=1$, $T=8$, so $\omega=\frac{\pi}{4}$ and $a=-\omega^2x=-\frac{\pi^2}{16}$ at $x=1$.
From the x–t graph, $A=1$, $T=8$ $\Rightarrow\omega=\frac{2\pi}{T}=\frac{\pi}{4}$ At $t=2$, $x=1$: $a=-\omega^2x$ $\Rightarrow a=-\frac{\pi^2}{16}\times1=-\frac{\pi^2}{16}\ m/s^2$ -
A simple pendulum oscillating in air has a period of $\sqrt3$ s. If it is completely immersed in non-viscous liquid, having density $\left(\dfrac{1}{4}\right)^{th}$ of the material of the bob, the new period will be:-$T_{air}=2\pi\sqrt{\dfrac{\ell}{g}}=\sqrt3$ sec. In the liquid, $g_{net}=g\left(1-\dfrac{\rho}{\sigma}\right)$ where $\rho$ is the density of the liquid and $\sigma$ is the density of the bob material. $T_{Liq}=2\pi\sqrt{\dfrac{\ell}{g_{net}}}=\dfrac{T_{air}}{\sqrt{1-\rho/\sigma}}=\dfrac{\sqrt3}{\sqrt{1-1/4}}=\dfrac{\sqrt3}{\sqrt3/2}=2$ sec.
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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :The longer pendulum (121 cm) has the larger time period. For them to be in phase again, the shorter one must complete one more oscillation: $n\,T_\ell = (n + 1)T_s$ $n \times 2\pi\sqrt{\frac{1.21}{g}} = (n + 1) \times 2\pi\sqrt{\frac{1}{g}}$ $1.1n = n + 1 \Rightarrow n = 10$ Number of oscillations of the shorter pendulum $= n + 1 = 11$
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Match List - I (x-y graphs) with List - II (Situations). Choose the correct answer.
Pick the matching item from Column II for each row.
Column I
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A. -
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B. -
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C. -
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D. -
Column II
- I. Total mechanical energy is conserved
- II. Bob of a pendulum is oscillating under negligible air friction
- III. Restoring force of a spring
- IV. Bob of a pendulum is oscillating along with air friction
Correct answer: A → IV, B → III, C → II, D → I
Graph (a) shows decreasing amplitude, i.e. damped oscillation: a pendulum oscillating with air friction (a-iv). Graph (b) is a straight line with negative slope, i.e. restoring force of a spring $F=-kx$ (b-iii). Graph (c) shows constant amplitude, i.e. undamped oscillation: pendulum with negligible air friction (c-ii). Graph (d) shows KE and PE curves whose sum is constant: total mechanical energy is conserved (d-i). -
A. -
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The restoring force of a spring with a block attached to the free end of the spring is represented byRestoring force of a spring is $F=-kx$, a straight line through the origin with negative slope.
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Identify the function which represents a non periodic motion.Periodic motion is represented by sine and cosine (harmonic) functions and their combinations. $e^{-\omega t}$ decreases monotonically and never repeats, so it is non-periodic.
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A body is executing simple harmonic motion with frequency $n$, the frequency of its potential energy is:Displacement in SHM of frequency n: $x = A\sin(2\pi nt)$ $U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2(2\pi nt) = \frac{1}{2}kA^2\left[\frac{1 - \cos(2\pi(2n)t)}{2}\right]$ So the frequency of potential energy = 2n.
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A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is:$F = kx \Rightarrow 10 = k(5 \times 10^{-2}) \Rightarrow k = 200$ N/m $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = \frac{2\pi}{10} = 0.628$ s
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The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:If $y = A\sin\omega t$, then $v = \frac{dy}{dt} = A\omega\cos\omega t$ $a = \frac{dv}{dt} = -A\omega^2\sin\omega t = A\omega^2\sin(\omega t + \pi)$ So the phase difference between displacement and acceleration is $\pi$ rad.
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Identify the function which represents a periodic motion.A periodic function repeats itself after a fixed interval of time. $\sin t+\cos t$ is a periodic function (period $2\pi$), while $e^{-t}$, $e^t$ and $\log_e(t)$ are not periodic.
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The displacement of a particle executing simple harmonic motion is given by $y=A_0+A\sin\omega t+B\cos\omega t$. Then the amplitude of its oscillation is given by:$y=A_0+A\sin\omega t+B\cos\omega t$ Shifting origin: $y'=y-A_0=A\sin\omega t+B\cos\omega t$ The two SHMs have a phase difference of $90^\circ$, so the resultant amplitude is $R=\sqrt{A^2+B^2+2AB\cos90^\circ}=\sqrt{A^2+B^2}$
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Average velocity of a particle executing SHM in one complete vibration is:In one complete vibration, the particle returns to its initial position, so displacement is zero. Average velocity $=\frac{\text{Displacement}}{\text{Time interval}}=\frac{y_f-y_i}{T}=\frac{0}{T}=0$
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The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the fig. $y$-projection of the radius vector of rotating particle $P$ is:
At $t=0$ the y-displacement is maximum, so the equation is a cosine function: $y=a\cos\omega t$
$\omega=\frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2}\ rad/s$, $a=3\ m$
$y(t)=3\cos\left(\frac{\pi t}{2}\right)$ -
The distance covered by a particle undergoing SHM in one time period is (amplitude = A):In one time period the particle goes from the mean position to +A, back to the mean position, to -A and back, covering a distance of 4A.
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