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Published on: 04/12/2019
Oscillation
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1.
The vertical motion of a huge piston in machine is simple harmonic with a frequency of 0.50 s-1. A block of 10 kg is placed on the piston. What is the maximum amplitude of the piston's SHM for the block and the piston to remain together?
2.
Can a motion be periodic but not oscillatory? If your answer is yes, give an example and if not explain why?
3.
Why does the time period of a swing not change when two persons sit on it instead of one?
4.
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillation is found to be 1.5 s. The radius of the disc is 15 cm.Determine the torsional spring constant of the wire.
This is a question based on torsion pendulum for which \(T=2\pi \sqrt { \frac { \quad }{ \alpha } } \) where I = moment of inertia of the disc about axis of rotation,\(\alpha \) = torsion constant which is restoring couple per unit twist.
5.
Justify the following statement
(i) The motion of an artificial satellite around the earth cannot be taken as SHM.
(ii) The time period of a simple pendulum will get doubled if its length is increased four times.
6.
Two pendulums of lengths 100 cm and 110.25 cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
7.
A particle is executing SHM according to the equation x = 5 sin \(\pi\)t where x is in cm. How long will the particle take to move from the position of equilibrium to the position of maximum displacement?
8.
A spring of force constant k has a mass M suspended from it. If the spring is cut into two halves, and the same mass is attached to one of the pieces, what will be the frequencies of oscillation of the mass?
9.
Two particles P and Q describe SHM of same amplitude a and frequency v along the same straight line. The maximum distance between two particles is √2 a. The phase difference between the particles is
zero
π/2
π/6
π/3
10.
A heavy brass sphere is hung from a spring and it executes vertical vibrations with period T. The sphere is now immersed in a non-viscous liquid with a density (1/10)th that of brass. When set into vertical vibrations with the sphere remaining inside liquid all the time, the time period will be
\(\sqrt{9\over 10T}\)
\(\sqrt{10\over 9T}\)
\(\sqrt{\left(9\over 10\right)r}\)
unchanged
11.
The following are the quantities associated with a body performing SHM.
1. The velocity of the body.
2. The accelerating of the body.
3. The accelerating force acting on the body.
Which of these quantities are exactly in phase with each other?
None of these
1 and 2 only
1 and 3 only
2 and 3 only
1, 2 and 3 only
12.
Masses in and 3m are attached to the two ends of a spring of constant k. If the system vibrates freely, the period of oscillation will be
\(\pi\sqrt{m\over k}\)
\(\pi\sqrt{3m\over 2k}\)
\(\pi\sqrt{3m\over k}\)
\(\pi\sqrt{4m\over 3k}\)
13.
A particle executing simple harmonic motion along y-axis has its motion described by the equation y = A sin (ωt) + B. The amplitude of the simple harmonic motion is
A
B
A+B
\(\sqrt{A+B}\)
14.
The length of a simple pendulum is increased by 44%. What is the percentage increase in its time period?
10%
20%
40%
44%
15.
A simple pendulum of frequency n is taken upto a certain height above the ground and then dropped along with its support so that it falls freely under gravity. The frequency of oscillations of the falling pendulum will
become greater than n
become zero
remain equal to n
become less than n
1.
As, \(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } \)
\(k=4\pi ^{ 2 }m{ v }^{ 2 }\)
For maximum displacement \({ y }_{ max }=A\)
Maximum restoring force,
F = - kA =- mg
or \(A=\frac { mg }{ k } =\frac { mg }{ 4{ \pi }^{ 2 }{ mv }^{ 2 } } =\frac { g }{ 4\pi ^{ 2 }{ v }^{ 2 } }\)
\( \\ =\frac { 9.8 }{ 4\times { (3.14) }^{ 2 }\times { (0.50) }^{ 2 } } =0.99m\)
2.
Yes, e.g., circular motion is periodic but not oscillatory.
3.
\(T=2\pi \sqrt { \frac { l }{ g } } \)so it does not depend upon the mass.
4.
Given mass of the disc m = 10 kg
Radius of the disc r = 15 cm = 0.15 m
T = 1,5 s
I is the moment of inertia of the disc about the axis of rotation which is perpendicular to the plane of the disc and passing through its centre.
\(I=\frac { 1 }{ 2 } { mr }^{ 2 }=\frac { 1 }{ 2 } \times (10)\times (0.15)^{ 2 }\)
= 0.1125 kg-m2
Time period, \(T=2\pi \sqrt { \frac { 1 }{ \alpha } } \)
\(\alpha =\frac { { 4\pi }^{ 2 }I }{ T^{ 2 } } =\frac { 4\times ({ 3.14) }^{ 2 }\times 0.1125 }{ { (1.5) }^{ 2 } } \)
\(\\ =1.972\ Nm/rad\)
5.
(i) The motion of an artificial satellite around the earth is periodic as it repeats after a regular interval of time. But it cannot be taken as SHM because it is not a to and fro motion about any fixed point that is, mean position.
(ii) Time period of simple pendulum,
\(T=2\pi \sqrt { \frac { l }{ g } } \) i.e.,\(T\alpha \sqrt { l } \)
Clearly, if the length is increased four times, the time period gets doubles.
6.
\(T=2\pi \sqrt{\frac{l}{g}}, l_1=100 cm\ ; l_2=110.25 cm\)
For smaller pendulum, \(T_1=2\pi \sqrt{\frac{100}{g}}\) -- (i)
For larger pendulum, \(T_1=2\pi \sqrt{\frac{110.25}{g}}\) -- (ii)
Let these pendulums oscillate in phase again if larger pendulum completes' n' oscillations. It means smaller pendulum must complete (n + 1) oscillations.
\(nT_2=(n+1)T_1\)
or \(\frac{n+1}{n}=\frac{T_2}{T_1}=\sqrt{\frac{110.25}{100}}=1.05\)
or \(1+\frac{1}{n}= 1.05 \ or \ \frac{1}{n}=0.05 = \frac{5}{100}=\frac{1}{20}\)
∴ n = 20.
Hence both pendulums will again oscillate in phase after 20 oscillations of the larger or 21 oscillations of the smaller pendulum.
7.
The displacement of the particle varies with time according to the equation.
x = 5 sin \(\pi\)t
Maximum displacement = amplitude = 5 cm
At time t = 0, x = 0 (equilibrium position). Hence time t taken by the particle to move from x = 0 to x = 5 cm is given by
5 = 5sin \(\pi\)t
or 1 = sin \(\pi\)t
or \(\pi\)t = \(\frac{\pi}{2}\)⇒t = 0.5 s.
8.
When the spring is cut into two equal halves, the force constant of each part will be doubled.
Therefore, the original frequency,
\(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ M } } \)
will becomes \(v'=\frac { 1 }{ 2\pi } \sqrt { \frac { 2k }{ M } } =\sqrt { 2 } v\)
9.
(a)
zero
10.
(b)
\(\sqrt{10\over 9T}\)
11.
(d)
2 and 3 only
12.
(c)
\(\pi\sqrt{3m\over k}\)
13.
(a)
A
14.
(b)
20%
15.
(b)
become zero
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