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Oscillations
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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?A Diagram showing constant $\omega(t)$ and decreasing $A(t)$
B Diagram showing $\omega(t)$ increasing with time and $A(t)$ decreasing with time
C Diagram showing $\omega(t)$ decreasing with time and $A(t)$ constant
D Diagram showing both $\omega(t)$ and $A(t)$ increasing with time
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:A $\frac{k_2}{k_1}$B $\frac{k_1}{k_2}$C $\sqrt{\frac{k_2}{k_1}}$D $\sqrt{\frac{k_1}{k_2}}$
$v_{max} = \omega A = \sqrt{k/m} A$. 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;A 5 cm, 1 sB 5 m, 1 sC 5 cm, 2 sD 5 m, 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:A $2\sqrt{3}$B 4C $\sqrt{3}$D $\sqrt{2}$
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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:
A $-\frac{\pi^2}{16}\ ms^{-2}$B $\frac{\pi^2}{8}\ ms^{-2}$C $-\frac{\pi^2}{8}\ ms^{-2}$D $\frac{\pi^2}{16}\ ms^{-2}$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 $t=2$, $x=1$: $a=-\omega^2x$ $\Rightarrow a=-\frac{\pi^2}{16}\times1=-\frac{\pi^2}{16}\ m/s^2$ -
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 :A 8B 11C 9D 10
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A body is executing simple harmonic motion with frequency $n$, the frequency of its potential energy is:A $2n$B $3n$C $4n$D $n$
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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:A 6.28 sB 3.14 sC 0.628 sD 0.0628 s
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The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:A $\frac{3\pi}{2}\ rad$B $\frac{\pi}{2}\ rad$C zeroD $\pi\ rad$
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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:A $A_0+\sqrt{A^2+B^2}$B $\sqrt{A^2+B^2}$C $\sqrt{A_0^2+(A+B)^2}$D $A+B$
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Average velocity of a particle executing SHM in one complete vibration is:A $\frac{A\omega}{2}$B $A\omega$C $\frac{A\omega^2}{2}$D Zero
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The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the fig. [add image] $y$-projection of the radius vector of rotating particle $P$ is:A $y(t)=-3\cos2\pi t$, where $y$ in mB $y(t)=4\sin\left(\frac{\pi t}{2}\right)$, where $y$ in mC $y(t)=3\cos\left(\frac{3\pi t}{2}\right)$, where $y$ in mD $y(t)=3\cos\left(\frac{\pi t}{2}\right)$, where $y$ in m