IISER Physics - Oscillations
Exam Duration: 45 Mins Total Questions : 30
In the case of simple pendulum, executing simple harmonic motion, the force supplying centripetal acceleration is
- (a)
\(mgcos\theta \)
- (b)
\(mgsin\theta \)
- (c)
\(-mgcos\theta \)
- (d)
\(-mgsin\theta \)
In the spring-mass system, if the mass of the system is doubled with spring constant halved, the natural frequency of longitudinal vibration
- (a)
is doubled
- (b)
is quadrupled
- (c)
is halved
- (d)
remains same
\(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } \)
\({ v }^{ \prime }=\frac { 1 }{ 2\pi } \sqrt { \frac { { v }^{ \prime } }{ m^{ \prime } } } =\frac { 1 }{ 2\pi } \sqrt { \frac { { k }/{ 2 } }{ 2m } } =\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ 4m } } =\frac { 1 }{ 2 } v\)
In the spring-mass system, the frequency of oscillation does not depend on
- (a)
the magnitude of displacement
- (b)
the magnitude of mass suspended
- (c)
the mass of the spring
- (d)
the magnitude of displacement and the mass of the spring
In spring-mass system, the spring is assumed an ideal.An ideal spring has negligible mass.
A mass M is attached to a spring whose upper end is fixed. The mass and stiffness k of the spring are m and k respectively. The natural frequency of the spring-mass system is
- (a)
\(\nu =\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ M+m } } \)
- (b)
\(r=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ M } } \)
- (c)
\(\nu =\frac { 1 }{ 2\pi } \sqrt { \frac { 3k }{ 3M+m } } \)
- (d)
\(r=\frac { 1 }{ 2\pi } \sqrt { \frac { 3k }{ M+3m } } \)
If m is the mass of the spring,
\(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ M+\frac { m }{ 3 } } } =\frac { 1 }{ 2\pi } \sqrt { \frac { 3k }{ 3M+m } } \)
where M is suspended mass.
For a simple pendulum in motion, if the effect of air resistance is taken into account, which parameter is constant of motion
- (a)
Energy
- (b)
Angluar momentum
- (c)
Restoring force
- (d)
Frequency of vibration
In a spring-mass system, of mass m and stiffness k, the ends of the spring are securely fixed and mass is attached to intermediate point of spring. The natural frequency of longitudinal vibration of the system.
- (a)
is minimum when the mass is attached to the mid-point of the spring
- (b)
is maximum when the mass is attached to the mid-point of the spring
- (c)
decreases as the distance from the bottom end whose mass is attached,decreases
- (d)
decreases as the distance from the top and where mass is attached, decreases
During the oscillations of a simple pendulum, the tension in the spring
- (a)
is greatest at its extreme position
- (b)
is zero at ats extreme position
- (c)
is greatest at its mean position
- (d)
is independent of mass
- (e)
is least at its mean position
Tension in the string at its mean position \(=mg+\frac { m{ \vartheta }^{ 2 } }{ r } \) where \(\vartheta \) is the velocity of mass at the mean position and r is the length of the simple pendulum. \(\vartheta \) is maximum at the mean position.
Tension in the string at its extreme position =\(mg\cos { \theta } \), where \(\vartheta \) is zero at its extreme position.
If the length of simple pendulum is quadrupled, the angular frequency will become
- (a)
one-fourth its previous value
- (b)
four times the previous value
- (c)
twice its previous value
- (d)
half its previous value
Angular frequency
\(\omega =\sqrt { \frac { g }{ l } } ,{ \omega }^{ \prime }=\sqrt { \frac { g }{ 4l } } =\frac { 1 }{ 2 } \sqrt { \frac { g }{ l } } =\frac { 1 }{ 2 } \omega \)
A disc of moment of inertia I suspended at the end of a slender wire executes 30 complete oscillations in one minute. To twist 10 degrees, a torque of 0.1 Nm is necessary. The value of I is (in kg m2)
- (a)
\(\frac { 1.8 }{ { \pi }^{ 3 } } \)
- (b)
\(\frac { 1.8 }{ { \pi }^{ 2 } } \)
- (c)
\(\frac { 1.8 }{ \pi } \)
- (d)
1.8
Time period of torsional oscillations, \(T=2\pi \sqrt { \frac { 1 }{ C } } \)
where C is the couple per unit twist (in radiation)
C = torque per unit radiation
\(=\frac { 0.1Nm }{ { 10 }^{ 0 } } \times \frac { { 180 }^{ 0 } }{ \pi rad } =\frac { 1.8 }{ \pi } .\frac { Nm }{ rad } \)
\(T=\frac { 60 }{ 30 } =2s\)
\(I=\frac { { T }^{ 2 } }{ 4{ \pi }^{ 2 } } C=\frac { 2\times 2 }{ 4{ \pi }^{ 2 } } .\frac { 1.8 }{ \pi } =\frac { 1.8 }{ { \pi }^{ 3 } } \)kg m2 = 0.058 kg m2
Consider the statements:
A. A body can have zero velocity and still be accelerating.
B. A body can have a northward velocity while experiencing a southward acceleration.
C. A body can have a constant speed and still have a varying velocity.
- (a)
A is correct only
- (b)
B is correct only
- (c)
C is correct only
- (d)
A,B,C are correct
Statement A is correct. Example is simple pendulum which at its extreme position experiences zero velocity but still be accelerating towards the mean position. Statement B is correct. Example is simple pendulum which always experiences acceleration towards mean position while moving away from it.
Statement C is correct. Example is body moving in a circular path.
A seconds pendulum is placed in an elevator at rest. When the elevator ascends with an acceleration \(4.9m{ s }^{ 2 }\),the pendulum will have time period (in s)
- (a)
2
- (b)
\(2\sqrt { 2 } \)
- (c)
\(2\sqrt { 3 } \)
- (d)
\(\sqrt { \frac { 8 }{ 3 } } \)
When an elevator ascends with an acceleration a, the effective acceleration becomes (g + a)
\(g+\frac { g }{ 2 } =\frac { 3g }{ 2 } \) where a = 4.9 ms-2 = \(\frac { g }{ 2 } \)
\(\therefore\) \({ T }_{ 2 }=2\pi \sqrt { \frac { 2l }{ 3g } } =\sqrt { \frac { 2 }{ 3 } } \left( 2\pi \sqrt { \frac { l }{ g } } \right) =\frac { \sqrt { 2 } }{ \sqrt { 3 } } \times 2\)
where the time period of a seconds pendulum is \(2\pi \sqrt { \frac { l }{ g } } \) which is 2 s.
Hence \({ T }_{ 2 }=\sqrt { \frac { 8 }{ 3 } s } \)
Two masses M and 16M are suspended from two identical springs. They are given small displecements in the same direction and at the same instant. They will be out of phase after mass M has completed
- (a)
one oscillation
- (b)
2 oscillations
- (c)
4 oscillations
- (d)
8 oscillations
\({ \omega }_{ 1 }=\sqrt { \frac { k }{ { M }_{ 1 } } } \) and \({ \omega }_{ 2 }=\sqrt { \frac { k }{ { M }_{ 2 } } } \)
\(\frac { { \nu }_{ 1 } }{ { { \nu } }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } =\sqrt { \frac { 16M }{ M } } =4\)
\({ \nu }_{ 1 }=4{ \nu }_{ 2 }\)
When mass M has completed 4 oscillation. mass 16 M, has completed 1 oscillation. Hence both masses are in the same phase. Mass. M will be out of phase after it completes 2 oscillations.
Simple harmonic motion is characterised as acceleration of a body is proportional to
- (a)
rate of change of velocity
- (b)
velocity
- (c)
mass
- (d)
NONE OF THE ABOVE
Simple harmonic motion is characterised as the acceleration of the body is proportional to displacement.
A helical spring of negligible mass is found to extend 0.25 mm under a mass of 1.5 kg. If mass 1.5 kg is replaced by mass of 60 kg, the system now will vibrate with a frequency of
- (a)
4.98 vibrations per second
- (b)
31.32 vibrations per second
- (c)
10.5 vibrations per second
- (d)
NONE OF THE ABOVE
1.5 kg-mass extends the spring by 0.25 mm.
60 kg - mass will extend it by
\(l=\frac { 0.25mm }{ 1.5kg } \times 60kg=10mm=10\times { 10 }^{ -3 }m\)
Required frequency \(\nu =\frac { 1 }{ 2\pi } \sqrt { \frac { g }{ l } } \)
\(=\frac { 1 }{ 2\pi } \sqrt { \frac { 9.8m{ s }^{ -2 } }{ 10\times { 10 }^{ -3 }m } } =4.98Hz\)
A mass m =2 kg is attached to a spring of stiffness \(8Nm^{ -1 }\).At time t=0 the mass is displaced to a position x=0.2 m and released from rest. The position x of the mass m is given by (in metre)
- (a)
x = 0.2 sin 2t
- (b)
\(x=0.2\quad sin\quad 4\pi t\)
- (c)
x = 0.2 cos 2t
- (d)
x = 2 cos 0.2 t
Angular frequency,
\(\omega =\sqrt { \frac { k }{ m } } =\sqrt { \frac { 8N{ m }^{ -1 } }{ 2kg } } \) = 2 rad s-1
Amplitude, a = 0.2 m
\(x=a\cos { 2\omega t } \)
\(=0.2\cos { 2t } \)
A particle moving along a straight line vibrates to and fro about the origin of a cartesian system. While passing through the origin it has
- (a)
zero potential energy and maximum kinetic energy
- (b)
minimum potential energy and maximum kinetic energy
- (c)
maximum potential energy and minimum kinetic energy
- (d)
minimum potential energy and minimum kinetic energy
A spring with zero relaxed length and spring constant k = 50 \({ Nm }^{ -1 }\) moves a block by contracting from a stretched length of 25 cm to a length of 5 cm. The block of mass m = 0.5 kg slides on a horizontal frictionless surface. The amount of work done on the block by the spring is
- (a)
2.188 J
- (b)
0.50 J
- (c)
1.500 J
- (d)
15 kJ
Work done
\(=\int _{ { \nu }_{ 1 } }^{ { { \nu } }_{ 2 } }{ -kx.dx } =\int _{ 0.05 }^{ 0.25 }{ -kx.dx } =\frac { 1 }{ 2 } x\left( { x }_{ 1 }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \)
\(=\frac { 1 }{ 2 } \times 50\left( { x }_{ 1 }+{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 2 } \right) \)
\(=\frac { 1 }{ 2 } \times 50\times 0.30\times 0.20=1.50J\)
A spring stretches by 3.0 cm from its released length when a force of 7.5 N is applied. A particle with a mass of 0.50 kg is attached to the free end of the spring, which is then compressed horizontally by 5.0 cm from its released length and released from rest at t=0. Then equation of motion of mass is
- (a)
\(x(t)=5.0sin\left( 22.36t+\frac { \pi }{ 2 } \right) \)
- (b)
\(x(t)=0.05sin\left( 22.36t+\frac { 3\pi }{ 2 } \right) \)
- (c)
\(x(t)=5.0sin\left( 22.36t+\frac { 3\pi }{ 2 } \right) \)
- (d)
\(x(t)=0.05sin\left( 22.36t+\frac { \pi }{ 2 } \right) \)
Spring constant,
\(k=\frac { -F }{ x } =-\frac { \left( -7.5N \right) }{ 0.03m } =\)250Nm-1
Angular frequency, \(\omega =\sqrt { \frac { k }{ m } } =\sqrt { \frac { 250N{ m }^{ -1 } }{ 0.50 } } =22.36\)rad s-1
Because spring is compressed, the initial displacement is negative (x0 < 0) and it is
x0 = -5.0 cm = 0.050 m
Amplitude = \(\sqrt { { x }_{ 0 }^{ 2 }+\frac { { \vartheta }_{ 0 }^{ 2 } }{ { w }^{ 2 } } } =\left| { x }_{ 0 } \right| =0.050\)m
where \({ \vartheta }_{ 0 }\) is initial velocity at t = 0. In fact x0 and \({ \vartheta }_{ 0 }\) are initial conditions at t = 0, and they are given by
\({ x }_{ o }=A\sin { { \phi }_{ o } } \)
\(\frac { { dx }_{ o } }{ dt } ={ \vartheta }_{ 0 }=a\omega \cos { { \phi }_{ o } } \) and \({ \vartheta }_{ 0 }=0\) at t=0.
Initial phase angle is \({ \phi }_{ o }=\tan ^{ -1 }{ \left( \frac { \omega { x }_{ o } }{ { \vartheta }_{ 0 } } \right) } \)
\(=\tan ^{ -1 }{ \left\{ \frac { \omega (-0.050) }{ 0 } \right\} } =\tan ^{ -1 }{ \left( -\infty \right) } =\frac { 3\pi }{ 2 } \)
Equation is \(x(t)=0.050\sin { \left( 22.36t+\frac { 3\pi }{ 2 } \right) } \)
A small bob with a mass of 0.20 kg hangs at rest from a massless string with a length of 1.40 m. At t=0 the bob is given a sharp horizontal blow that delivers an impulse, \(J=\int { F\quad dt=0.15\quad Ns } \) due to which it gets an angular displacement \(\theta \). The equation of motion of the bob (in radian) is
- (a)
\(\theta (t)=0.202sin\quad 5.30t\)
- (b)
\(\theta (t)=0.404sin\quad 1.37t\)
- (c)
\(\theta (t)=0.202sin\quad 2.65t\)
- (d)
\(\theta (t)=0.303sin\quad 2.02t\)
A block with a mass M = 0.50 kg is suspended at rest from a spring with spring constant k=200 \(N{ m }^{ -1 }\). A blob of putty (m=0.30 kg) is dropped onto the block from a height of 10 cm; the putty slicks to the block. The period of the ensuring oscillations is
- (a)
1.2 s
- (b)
0.397 s
- (c)
0.252 s
- (d)
4.2 s
\({ \omega }=\sqrt { \frac { k }{ m } } \)
\({ \omega }^{ \prime }=\sqrt { \frac { k }{ M+m } } =\frac { 2\pi }{ T } { T }^{ \prime }=2\pi \sqrt { \frac { M+m }{ k } } \)
\(=2\pi \sqrt { \left( \frac { 0.50+0.30 }{ 200 } \right) } =0.397s\)
A block with a mass M = 0.50 kg is suspended at rest from a spring with spring constant k=200 Nm−1. A blob of putty (m=0.30 kg) is dropped onto the block from a height of 10 cm; the putty slicks to the block. The total energy of the oscillating system is
- (a)
0.132 J
- (b)
1.32 J
- (c)
0.120 J
- (d)
13.2 J
First, calculate amplitude which is found to be 0.0363 m
Total energy = \(\frac { 1 }{ 2 } k{ A }^{ 2 }=\frac { 1 }{ 2 } \)(200)(0.0363)2 = 0.1318J
The displacement y of particle executing periodic motion is given by \(y=4{ cos }^{ 2 }(t/2)sin(1000t)\) this expression may be considered to be a result of the superposition of how many independent harmonic motions
- (a)
Five
- (b)
Two
- (c)
Three
- (d)
Four
Given \(y=2\cos ^{ 2 }{ \frac { t }{ 2 } } \sin { 1000t } \)
\(=2\cos { \frac { t }{ 2 } } 2\cos { \frac { t }{ 2 } } \sin { 1000t } \)
\(=2\cos { \frac { t }{ 2 } } \left[ \sin { \left( 1000t+\frac { t }{ 2 } \right) -\sin { \left( 1000t-\frac { t }{ 2 } \right) } } \right] \)
It is the superposition of three independent harmonic motions.
A system exhibiting S.H.M. must possess
- (a)
elasticity as well as inertia
- (b)
elasticity,inertia and an external force
- (c)
elasticity only
- (d)
inertia only
A particle executing S.H.M. of amplitude 4 cm and time period T = 4s. The time taken by it to move from positive extreme position to half the amplitude is
- (a)
\(\sqrt { { 3 }/{ 2 } } s\)
- (b)
1 s
- (c)
1/3 s
- (d)
2/3 s
It is given that time is noted from the extreme position. This means that displacement (measured from the mean position) is maximum at time t = 0. Hence, equation of SHM is
y = a \(\cos { \omega t } \)
When \(y=\frac { a }{ z } \), \(\frac { a }{ 2 } =a\cos { \omega t } \)
\(\cos { \omega t } =\frac { 1 }{ 2 } \)
\(\omega t=\frac { \pi }{ 3 } \)
\(\frac { 2\pi }{ T } t=\frac { \pi }{ 3 } \)
\(t=\frac { T }{ 6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } s\)
Identify the correct statement among the following :
- (a)
A simple pendulum with a bob of mass M swings with an angular amplitude of 40 °. When its angular amplitude is 20 °, then the tension in the string is Mg cos 20 °.
- (b)
The greater the mass of a pendulum bob, the shorter is its frequency of oscillation.
- (c)
The fractional change in the time period of a pendulum on changing the temperature is independent of the length of the pendulum.
- (d)
As the length of a simple pendulum is increased, the maximum velocity of its bob during its oscillation will also increase.
Option (a): For amplitude of bob equal to 200.
T-mg\(\cos { \theta } \) = \(\frac { m{ v }^{ 2 } }{ r } \) gives \(T\neq mg\cos { \theta } \)
Option(b): \(\omega =\sqrt { \frac { g }{ l } } \). As l is increased, frequency decreases.
option (d): \(\omega =\sqrt { \frac { g }{ l } } \)
\({ \vartheta }_{ max }=a\omega =a\sqrt { \frac { g }{ l } } \)
As l increases, \({ \vartheta }_{ max }\) decreases
Option(c): Let \(\alpha\)= temperature coefficient of linear expansion of suspension string. l1 = length of simple pendulum at ternperature t0C, then l1 = l(1+\(\alpha\)t)
Time period T = \(2\pi \sqrt { \frac { l }{ g } } \) and \({ T }_{ 1 }=2\pi \sqrt { \frac { { l }_{ 1 } }{ g } } \)=\(\sqrt { \frac { l\left( l+\alpha t \right) }{ g } } \)
\(\frac { { T }_{ 1 } }{ T } =\sqrt { 1+\alpha t } \) . This is independent of l
A particle of mass m is executing S.H.M. about the origin along x-axis. The P.E. U(x) = kx3 where k is positive constant. If the amplitude of oscillation is a, then time period T is proportional to
- (a)
\(\frac { 1 }{ \sqrt { a } } \)
- (b)
\(\sqrt { a } \)
- (c)
\({ a }^{ 3/2 }\)
- (d)
independent of a
In SHM, the time period is independent of amplitude.
The angular velocity, and the amplitude of a simple pendulum is \('\omega '\)and 'a'. At a displacement x from the mean position, the kinetic energy is T and the potential energy is V. Then the ratio of T to V is
- (a)
\(\frac { { x }^{ 2 }{ \omega }^{ 2 } }{ { A }^{ 2 }-{ x }^{ 2 }{ \omega }^{ 2 } } \)
- (b)
\(\frac { { x }^{ 2 } }{ { A }^{ 2 }-{ x }^{ 2 } } \)
- (c)
\(\frac { { { A }^{ 2 }-x }^{ 2 }{ \omega }^{ 2 } }{ { x }^{ 2 }-{ x }^{ 2 }{ \omega }^{ 2 } } \)
- (d)
\(\frac { { A }^{ 2 }-{ x }^{ 2 } }{ { x }^{ 2 } } \)
P.E. is \(v=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ x }^{ 2 }\)
K.E is \(T=\frac { 1 }{ 2 } m{ \omega }^{ 2 }\left( { a }^{ 2 }-{ x }^{ 2 } \right) \)
\(\frac { T }{ V } =\frac { \left( { a }^{ 2 }-{ x }^{ 2 } \right) }{ { x }^{ 2 } } \)
A test tube of cross-section 'a' has some lead shots at the bottom so that total mass of the tube is 'm'. It floats vertically in a liquid of density d. It is then pushed through a distance into the liquid and released If l is length of the tube dipping initially, in the liquid, the time period of oscillation is
- (a)
\(T=2\pi \sqrt { \frac { l }{ g } } \)
- (b)
\(T=2\pi \sqrt { \frac { m }{ g } } \)
- (c)
\(T=2\pi \sqrt { \frac { md }{ g } } \)
- (d)
\(T=2\pi \sqrt { \frac { md }{ ag } } \)
By the law of floatation. mg = (Al) \(\rho \)g
or mass, m = Al\(\rho \)
Restoring force, F = (Ay)\(\rho \) g
where y is the distance to which the tube is depressed
Acceleration, a = \(\frac { -Ay\rho g }{ Al\rho } =\frac { -g }{ l } y\)
Comparing with equation of SHM, a = \(-{ \omega }^{ 2 }y\)
\({ \omega }^{ 2 }=\frac { g }{ l } \) or \({ \omega }=\sqrt { \frac { g }{ l } } \) or \(T=2\pi \sqrt { \frac { l }{ g } } \)
When the displacement is half of the amplitude, then what fraction of the total energy of a simple harmonic oscillator is kinetic?
- (a)
2/7th
- (b)
3/4th
- (c)
2/9th
- (d)
5/7th
K.E. is \({ E }_{ k }=\frac { 1 }{ 2 } m{ \omega }^{ 2 }\left( { a }^{ 2 }-{ y }^{ 2 } \right) \)
\(=\frac { 1 }{ 2 } m{ \omega }^{ 2 }\left( { a }^{ 2 }-\frac { { a }^{ 2 } }{ 4 } \right) =\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }\left( \frac { 3 }{ 4 } \right) \)
Total energy = Maximum K.E. = \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }\)
Ek is 3/4 th of total energy
For a particle executing simple harmonic motion, which of the following statements is correct?
- (a)
Total energy of the particle always remains the same
- (b)
Restoring force is always directed towards fixed point
- (c)
Restoring force is maximum at the equilibrium position
- (d)
Acceleration of the particle is maximum at the extreme position
Acceleration, a = -\({ \omega }^{ 2 }\)A at the extreme position when A' is the amplitude of SHM.