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The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10-2 cos πt metres. The time at which the maximum speed first occurs
0.5 s
0.75 s
0.125 s
0.325 s
A.
0.5 s
x = 2 x 10-2 cos πt
v = 0.02 π sinπt
v is maximum at t = 1/2 = 0.5 sec
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
at the highest position of the platform
at the mean position of the platform
for an amplitude of g/ω2
for an amplitude of g2/ ω2
C.
for an amplitude of g/ω2
Aω2 = g
⇒ A = g/ω2
Two simple harmonic motions are represented by the equation y1 = 0.1 and y2 = 0.1 cosπt. The phase difference of the velocity of particle 1 w.r.t. the velocity of the particle 2 is
−π/6
π/3
−π/3
π/6
A.
−π/6
Phase difference (φ) = 99πt + π/3 −π/2 at
t = 0 φ = −π/6
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillation bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would
first increase and then decrease to the original value.
first decreased then increase to the original value.
remain unchanged.
increase towards a saturation value.
A.
first increase and then decrease to the original value.
First CM goes down and then comes to its initial position.
The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in the air. Neglecting frictional force of water and given that the density of the bob is (4/3) x 1000 ms-1 . What relationship between t and t0 is true?
t = t0
t = t0/2
t = 2t0
t = 4t0
C.
t = 2t0
The time period of simple pendulum in air
being the length of simple pendulum, In water, effective weight of bob
w' = weight of bob in air - upthrust
⇒ ρVgeff = mg-m'g = ρVg-ρ'Vg = (ρ-ρ') where ρ = density of bob, ρ′ = density of water
A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is t, then
T = t1 + t2
B.
Time period of spring
k, being the force constant of spring. For first spring and for second spring
we have,
The effective force constant in their series combination is
The total energy of particle, executing simple harmonic motion is
∝ x
∝ x2
independent of x
∝ x1/2
C.
independent of x
In simple harmonic motion when a particle is displaced to a position from its mean position, then its kinetic energy gets converted into potential energy and vice-versa. Hence, total energy of a particle remains constant or the total energy in simple harmonic motion does not depend on displacement x.
A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω0. An external force F(t) proportional to cosωt (ω≠ω0) is applied to the oscillator. The time displacement of the oscillator will be proportional to
B.
Initial angular velocity of particle = ω0 and at any instant t, angular velocity = ω Therefore, for a displacement x, the resultant acceleration
Now, equation of simple harmonic motion
x = A sin (ωt + φ) .......... (iv)
at t = 0 ; x = A
∴ A = A sin( 0 + φ )
⇒ φ =π/2 ..........(v)
Hence, from equations (iii) and (v), we finally get
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In forced oscillation of a particle the amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω2 of the force, then
ω1 = ω2
ω1 > ω2
ω1 < ω2 when damping is small and ω1 > ω2 when damping is large
ω1 < ω2
A.
ω1 = ω2
For amplitude of oscillation and energy to be maximum, frequency of force must be equal to the initial frequency and this is only possible in case of resonance. In resonance state ω1 = ω2
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively
g,g/3
g/3, g
g,g
g/3, g/3
B.
g/3, g
Before the string is cut
kx = T + 3mg ...(1)
T = mg .. (2)
⇒
kx = 4mg
After the string is cut, T = 0
A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27ºC two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27ºC is
330 m/s
339 m/s
300 m/s
350 m/s
B.
339 m/s
Two successive resonance is produced at 20 cm and 73 cm of column length
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is
2πs
π s
1s
2s
B.
π s
Acceleration, a = 20 m/s2 , and displacement, y = 5
A particular of mass m is executing oscillation about the origin on X- axis. Its potential energy is V(x) = K |x|3. Where K is a positive constant. If the amplitude of oscillation is a, then its time period T is proportional to
a
a3/2
A.
Given, V = kx3
Total energy (E) = Maximum potential energy
= ka3
From conservation of energy
The displacement of a particle along the x-axis is given by x = a sin2 ωt. The motion of the particle corresponds to
Simple harmonic motion of frequency ω/π
Simple harmonic motion of frequency 3ω/2π
non-simple harmonic motion
Simple harmonic motion of frequency ω/2π
C.
non-simple harmonic motion
Given X = a sin2 ωt
Here, a is not directly proportional to (-x) which is a condition for SHM.
A simple pendulum of length l has a maximum angular displacement θ. The maximum kinetic energy of the bob is
mgl (1-cosθ)
0.5 mgl
mgl
0.5 mgl
A.
mgl (1-cosθ)
Here length of the pendulum = l
Maximum angular displacement =θ
Mass of the bob = m
Now height of the bob at maximum angular displacement is
h = l - l (1-cosθ)
Also at the end of displacement kinetic energy of the bob = potential energy of the bob
mgh = mgl (1-cosθ)
hence kinetic energy of the bob = mgl (1-cosθ)
Equations of motion in the same direction are given by
The amplitude of the medium particle will be
C.
The given equations of motion are
If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period becomes :
3 T
3/2T
4T
2T
A.
3 T
The time for one complete cycle, a left swing, is called the period. The period depends on length of the period. The period depends on the length of the pendulum and also to a slight degree on the amplitude, the width of the pendulum's swing.
A pendulum is undergoing SHM with frequency f What is the frequency of its kinetic energy ?
2 f
3 f
4 f
B.
2 f
Suppose A particle in s.H.M follows the equation
Two equal -ve charges -q are fixed at the point (O, a) and (O, a) on the y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge will :
execute SHM about the origin
move to the origin and remain at rest
move to infinity
execute oscillatory but not SHM
D.
execute oscillatory but not SHM
Component of force on charge +Q at P, along the x-axis,
Which is directly proportional to x. So, motion is an oscillatory but not SHM.
The circular motion of a particle with constant speed is
simple harmonic but not periodic
periodic and simple harmonic
neither periodic nor simple harmonic
periodic but not simple harmonic
D.
periodic but not simple harmonic
In a circular motion particle repeats after equal intervals of time. So, particle motion on a circular path is periodic but not simple harmonic as it does not execute to and fro motion about a fixed point.
A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is
3 Hz
2 Hz
4 Hz
1 Hz
D.
1 Hz
The maximum speed of a particle executing SHM is
substituting the given values, we have
A particle executes simple harmonic oscillation with an amplitude . The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
C.
Let displacement equation of particle executing SHM is
As the particle travels half of the amplitude from the equilibrium position, so
Hence, the particle travels half of the amplitude from the equilibrium in .
Given that , where y and x are measured in metre. Which of the following statements is true?
The unit of λ is same as that of x and A
The unit of λ is same as that of x but not of A
The unit of c is same as that of
The unit of ( ct - x ) is same as that of
A.
The unit of λ is same as that of x and A
Here is dimensionless
Hence, is also dimensionless and unit of ct is the same as that of x. Therefore unit of λ is same as that of x.
Also unit of y is same as that of A, which is also the unit of x.
Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has
completed x oscillations, where k is
1
3
2
5
C.
2
Time period of simple pendulum
For 1st pendulum T1 =
=
=
For IInd pendulum T2 =
= 2
= 2
Suppose the time will be in same phase is t then
=
=
Hence the number of oscillation of simple pendulum of shorter wavelength
(x) =
(x) = 2
A mass of 1 kg is suspended from a spring of force constant 400 N, executing SHM total energy of the body is 2J, then maximum acceleration of the spring will be
4 m/s2
40 m/s2
200 m/s2
400 m/s2
B.
40 m/s2
Oscillations due to spring
T = 2
Where T is the time period
k is spring constant
m is mass attached to spring
T =
= 2
=
Frequency n =
ω = 2π n
= 2π ×
ω = 20
Total energy
E =
2 =
r =
Acceleration = ω2 r
= 400
= 40 m/s2
Vibrations of rope tied by two rigid ends shown by equation y = cos 2πt sin2πx, then minimum length of the rope will be
1 m
5 m
2πm
B.
Equation of vibration
y = 1( cos2 sin2 )
= 1 cos2 sin
Comparing this equation with
y =
Hence fro above equation
a = 1
Minimum length of string
=
=
Mininmum length of string
=
The angular amplitude of a simple pendulum is θ0. The maximum tension in its string will be
mg (1 - θ0)
mg (1 + θ0 )
D.
The simple pendulum at angular amplitude θo is shown in the figure
Tmax = ....(i)
When bob of pendulum comes from A to B, it covers a vertical distance h.
∴
⇒ h = l ( l - cos θ ) .....(ii)
Also during A to B, potential energy of bob converts into kinetic energy
∴ mgh =
∴ v = ....... (iii)
Thus, using Eqs. (i), (ii) and (iii) we obtain
Tmax = mg +
= mg + 2mg
(By using binomial theorem)
= mg
A particle is subjected to two simple harmonic motions along X-axis while other is along a line making angle 45° with the X-axis. The two motions are given by x = x0 sin ωt and s =s0 sin ωt. The amplitude of resultant motion is
x0 + s0 + 2x0s0
D.
Two SHMs are separated at angle 45° So, resultant amplitude is
AR =
=
Since sin 45o = ⇒ sin2 45o =
cos 45o = ⇒ cos2 45o =
AR =
⇒ AR =
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A pan with a set of weights is attached to a light spring. The period of vertical oscillation is 0.5s. When some additional weights are put in pan, then the period of oscillations increases by 0.1 s. The extension caused by the additional weight is
5.5 cm
3.8 cm
2.7 cm
1.3 cm
C.
2.7 cm
Given:- T = 0.5 s
We know that,
The time period for oscillation is
T =
k = It is spring constant
m = mass
⇒ T = 0.5
⇒ 0.5 =
⇒
On squaring on both sides, we get
⇒ ....(i)
⇒ T + 0.1 =
⇒ 0.5 + 0.1 = 2
⇒
⇒ =
⇒ + using eq.(i)
⇒
⇒
⇒ ....(ii)
m' g = kx
x = g
⇒ = × 10
⇒ x = 0.0279
⇒ x = 2.79 × 10-2 m
× x = 2.7 cm
A particle is describing simple harmonic motion. If its velocities are v1 and v2 when the displacements from the mean position are y1 and y2 respectively, then its time period is
D.
In simple harmonic motion,
velocity
v =
∴ v1 =
⇒ .....(i)
and
⇒ ....(ii)
Solving equations (i) and (ii), we get
Time period is given by,
⇒ T =
⇒ T =
Assertion: A particle in x-y plane is related by x = sin t and y = (l - cos t), where and ω are constants, then the particle will have parabolic motion.
Reason: A particle under the influence of two perpendicular velocities has parabolic motion.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
D.
If both assertion and reason are false.
Both assertion and result are false.
A body of mass 60 kg is suspended by means of three strings P Q and R as shown in the figure is in equilibrium. The tension in the string P is
130.9 g N
60 g N
50 g N
103.9 g N
D.
103.9 g N
R cos 60o = Q = 60 g
⇒ R =
P = R sin60o
= 120 g ×
⇒ P = 103.9 g N
The angular amplitude of simple pendulum is θo. The maximum tension in its string will be
mg ( 1 θ0 )
mg ( 1 + cosθ0 )
mg ( 1 cos )
mg ( 1 + cos )
D.
mg ( 1 + cos )
Maximum tension in the string is
Tmax = mg +
= mg + ( 1 - cosθ0 )
= mg + 2mg ( since θ0 is small )
Tmax = mg ( 1 + θ0 )
The displacement of a particle executing SHM is given by y = 0.25 sin200t cm. The maximum speed of the particle is
200 cm s-1
100 cm s-1
50 cm s-1
5.25 cm s-1
C.
50 cm s-1
Given:- displacement of particle executing SHM
y = 0.25 sin 200t
speed =
= 0.25 × 200 cos 200t
i.e cos 200t = 1
Max speed= 0.25 x 200
= 50 cm s-1
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, then the time period becomes . The ratio of is
9/16
5/4
25/16
4/5
A.
9/16
The time period of oscillation is
T = 2
M = mass attached to spring
and T1 = 2
⇒
⇒
⇒ 1
⇒
A wave is represented by the equation y= 0.5 sin(10t x) metre It is a travelling wave propagating along +x direction with velocity
10 m s-1
20 m s-1
5 m s-1
None of these
A.
10 m s-1
Compare the given eqution y = 0.5 sin (10t x) with the standard form
y = r sin
= 10
= 1
⇒
=
⇒ v = 10 m/s
The frequency of oscillations of a mass m connected horizontally by a spring of spring constant k is 4 Hz. When the spring is replaced by two identical spring as shown in figure. Then the effective frequency is
4
1.5
1.31
2
D.
2
If one spring is connected then frequency = 4Hz
4 = ....(i)
If two springs are connected in series, then frequency
v = ....(ii)
From equation (i) and (ii)
v = 2 Hz
Assertion: The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola.
Reason: Potential energy and kinetic energy do not vary linearly with position.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
B.
If both assertion and reason are true but reason is not the correct explanation of assertion.
Both assertion and reason are true but reason is not the correct explanation of assertion.
P. E = m ω2 y2
and K.E = m ω2 ( α2 - y2 )
The variation of both potential energy and kinetic energy is proportional to y2.
Therefore displacement versus P.E. or K.E. graph is a parabola ( because parabolic equation is z = 4au2 )
Motion of an oscillating liquid in a U tube is
periodic but not simple harmonic.
non-periodic.
simple harmonic and time period is independent of the density of the liquid.
simple harmonic and time period is directly proportional to the density of the liquid.
C.
simple harmonic and time period is independent of the density of the liquid.
Motion of an oscillating liquid column in a U tube is SHM with period
T =
where h is the height of liquid column in one arm of U tube in equilibrium position of
liquid. Therefore, T is independent of density of liquid.
The displacement x of a particle varies with time t as, where a, b α and β are positive constants. The velocity of the particle will
go on decreasing with time
be independent of
drop to zero when
go on increasing with time
D.
go on increasing with time
The value of term decreases and of term increases with increase in time. As a result, velocity goes on increasing with time.
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