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Oscillations And Wave Mcq
Quiz 3
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Q.1
If a spring has time period T and is cut into n equal parts, then the time period of each part will be [ AIEEE 2002]
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a) T√n
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b) T / √n
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c) nT
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d) T
Explanation
Let the spring constant of the original spring be k when cut in n equal parts force constant of each part become nk If T is periodic time before cutting T=2π √(m/k) new periodic time T'=2π √(m/n) T'=T/√n Answer: (b)
Q.2
A child swinging on a swing in sitting position, stand up, then the time period of the swing will ..[ AIEEE 2002]
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a) increase
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b) decrease
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c) remains same
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d) increases of the child is long and decreases if the child is short
Explanation
the time period of swing T ∝√l here l is the between the point of suspension and the centre of mass of the child as child stands up centre of mass moves up reducing l thus time period will be reduced Answer: (b)
Q.3
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 SHM of period T. If the mass is increased by m, the time period becomes 5T/Then the ratio of m/M is ..[ AIEEE 2003]
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a) 3/5
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b) 25/9
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c) 16/9
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d) 5/3
Explanation
Answer: (c)
Q.4
Two particles A and B of equal masses are suspended from two massless springs of spring constant k1 and k2 respectively. If the maximum velocities, during oscillation are equal, the ratio of amplitude of A and B is [ AIEEE 2003]
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a) √(k1 / k2)
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b) k2 / k1
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c) √(k2 / k1)
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d) k1 / k2
Explanation
Maximum velocity during SHM = Aω , but k = mω2 ∴ ω = √(k/m) ∴ Maximum velocity = A√(k/m) Here the maximum velocity is same and m is also same A1√k1 = A2√k2 ∴ A1 / A2 = √(k2 / k1) Answer:(c)
Q.5
The length of a simple pendulum executing simple harmonic motion is increased by 21%. the percentage increase in the time period of the pendulum of increased length is ..[ AIEEE 2003]
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a) 11%
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b) 21%
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c) 42%
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d) 10%
Explanation
given l' = 1.21l Thus T' = 2π √(1.21l/g) % increase = Answer: (d)
Q.6
The displacement of particle varies according to the relation x=4(cosπt+sinπt). The amplitude of the particle is ..[ AIEEE 2003]
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a)-4
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b) 4
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c)4√2
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d)8
Explanation
given y=4(cosπt+sinπt)Multiplying and dividing by √2 we get x=√2 ×4( sinπt/ √2 + cosπt/√2) x=4√2 ( sinπt cos45 + cosπtsin45) x=4√2 sin(πt +45) on comparing it with x=A sin(œt +φ) we get A=4√2Answer: (c)
Q.7
A body executes simple harmonic motion. The potential energy (P.E.) , kinetic energy(K.E.) and total energy (T.E) are measured as a function of displacement x. Which of the following statement is true? [ AIEEE 2003]
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a) K.E. is maximum when x=0
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b) T.E is zero when x=0
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c)K.E. is maximum when x is maximum
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d)P.E. is maximum when x=0
Explanation
K.E=½ mω2 ( a2 - x2) When x=0, K.E is maximum Answer: (a)
Q.8
The bob of simple pendulum executes simple harmonic oscillation in water with period t, while the period of oscillation of bob in air is to. Neglect frictional force of water and given that the density of boob is (4/3)×103 kg/mWhat relationship between t and to is true [ AIEEE 2004]
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a) t=2to
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b) t=to/2
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c)t=to
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d)t=4to
Explanation
When bob is in water buoyant force act on it as a result effective gravitational acceleration is lesser in water than air Wt of bob in water=weight in air - Buoyant forcevρg'=vρg - vρoghere ρ density of boob , ρo density of water, v is volume of boobg'=g( ρ - ρo) / ρ Formula for periodic time of simple pendulum Answer:(a)
Q.9
A particle at the end of a spring execute S.H.M. with a period t1, while the corresponding period for another spring is tIf the period of oscillation with the to springs in series is T then ..[ AIEEE 2004]
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a) T-1=t1-1 + t2-1
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b) T2=t12 + t22
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c) T=t1 + t2
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d) T-2=t1-2 + t2-2
Explanation
For first spring t1=2π √( m/k1) For second spring t2=2π √( m/k2)When springs are connected in series effective spring constant k eff=k1k2 / k1 + k2 But k1=4π2m / t12 k2=4π2m / t22 substituting values of k1 and k2 in equation of T we get Answer: (b)
Q.10
The total energy of particle, executing simple harmonic motion is [AIEEE 2004]x is the displacement for equilibrium position
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a)in dependant of x
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b) ∝ x2
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c)∝ x
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d)∝ x 1/2
Explanation
At any instant total energy is ½ kAo2=constant Here Ao is amplitude of oscillation. Thus total energy is in dependant of displacement xAnswer: (a)
Q.11
A particle of mass m is attached to a spring ( of spring constant k) and has a nature frequency ωo. An external force F(t) proportional to cosωt ( ω ≠ ωo) is applied to the oscillator. the time displacement of the oscillator will be proportional to ... [ AIEEE 2004]
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a)
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b)
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c)
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d)
Explanation
Equation for displacement is given by x=A sin ( ωt + φ) Where A is here damping coefficient is considered to be zero ∴ option b is correctAnswer: (b)
Q.12
In force oscillation of a particle the amplitude is maximum for frequency ω1 of the force while energy is maximum for frequency ω2 of the force then [ AIEEE 2004]
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a) ω1 < ω2 when damping is small and ω1 > ω2 when damping is large
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b)ω1 > ω2
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c)ω1=ω2
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d)ω1 < ω2
Explanation
Since energy ∝ ( Amplitude ) 2 , the maximum energy will be same for both frequency Answer:(c)
Q.13
Two simple harmonic motions are represented by equations The phase difference of the velocity of particle 1 with respect to the velocity of the particle 2 is .. [ AIEEE 2005]
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a) π / 3
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b) -π/6
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c) π/6
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d) -π/3
Explanation
Taking derivative of both equations Phase difference=π/3 - π/2=- π/6 Answer: (b)
Q.14
The function sin2(ωt) represents [ AIEEE 2005]
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a)a periodic, but not simple harmonic function with period π/ω
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b) a periodic, but not simple harmonic motion with period 2π/ω
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c)a simple harmonic motion with a period π/ω
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d)a simple harmonic motion with a period 2π / ω
Explanation
sin2ωt is periodic with period π/ ω For function to SHM d2y/ dt2=-y is a conditionclearly second derivative is not proportional to -y. Hence not SHMAnswer: (c)
Q.15
The bob of a simple pendulum is spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of the oscillation would [ AIEEE 2005]
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a) first decreases and then increase to the original value
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b) first increase and then decrease to the original value
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c)increase towards a saturation value
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d)remains unchanged
Explanation
Periodic time of simple pendulum depends on the length of pendulumWhen water starts to come out centre of mass of bob shifts to lower position thus length increases and period of oscillation increases. When water goes out completely. centre of mass regain its original position thus option "b" is correctAnswer: (b)
Q.16
If a simple harmonic motion is represented by d2x/dt2 + αx=0, its time period is [ AIEEE 2005]
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a) 2π /√α
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b) 2π /α
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c)2π √α
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d)2πα
Explanation
Comparing given equation with standard equation for SHM ω2=α∴ (2π /T)2=α T=2π /√α Answer:(a)
Q.17
The maximum velocity of a particle executing simple harmonic motion with an amplitude 7mm is 4.4m/s. The period of oscillation is ..[ AIEEE 2006]
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a) 0.01 s
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b) 10 s
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c) 0.1s
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d) 100s
Explanation
Maximum velocity vmax=aω vmax=a(2π /T) T=2πa / vmax T=2×3.14×7×10-3 / 4.4=0.01s Answer: (a)
Q.18
Starting from the origin a body oscillates simple harmonically with a period of 2s. After what time will its kinetic energy be 75%, of the total energy? [ AIEEE 2006]
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a)(1/6) s
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b) (1/4) s
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c)(1/3) s
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d)(1/12)s
Explanation
K.E of a body undergoing SHM is given by K.E=½ ma2ω2cos2ωt T.E.=½ma2ω2 Given K.E=075T.E 0.75=cos2ωt ∴ t=π / (6×ω)t=(π×2) / (6×ω)=(1/6) sAnswer: (a)
Q.19
Two springs, of force constants k1 and k2 are connected top a mass m as shown. the frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes .. [ AIEEE 2007]
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a) 2f
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b) f/2
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c)f/4
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d)4f
Explanation
Two springs are parallel ∴ effective spring constant k=k1 + k2Now, frequency of oscillation is given by When both k1 and k2 are made four times their original value, the new frequency is given by Answer: (a)
Q.20
A particle of mass m executes simple harmonic motion with amplitude a and frequency ν. the average kinetic energy during its motion from the position of equilibrium to the end is ..[ AIEEE 2007]
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a)2π2ma2v2
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b) π2ma2v2
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c)(1/4)ma2v2
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d)4π2ma2v2
Explanation
The kinetic energy of a particle executing S.H.M. is given by K=½ ma2ω2sin2ωt Here ω=angular frequency m=mass of particle a=amplitudet=timeaverage K.E=
=< ½ ma2ω2sin2ωt >
=½ ma2ω2
=½ ma2ω2(1/2)
=½ ma2ω2(1/2)
=(1/4) ma2ω2 since ω=2πν
=(1/4) ma2(2πν)2
=π2ma2v2 Answer:(b)
Q.21
The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10-2cosπt meter. the time at which the maximum speed first occurs is .. [ AIEEE 2007]
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a) 0.25s
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b) 0.5s
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c) 0.75s
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d) 0.125s
Explanation
Here, x=2×10-2cosπt speed is given by v=dx/dt=2×10-2πsinπt For the first time, the speed to be maximum sinπt=1 or sinπt=sinπ/2 ∴ πt=π/2 or t=1/2=0.5sec Answer: (b)
Q.22
A point mass oscillates along x-axis is according to the law x=xocos(ω- π/4). If the acceleration of the particle is written as a=Acos(ωt-δ), then ...[ AIEEE 2007]
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a)A=xoω2, δ=3π/4
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b) A=xo, δ=-π/4
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c)A=xoω2, δ=π/4
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d)A=xoω2, δ=-π/4
Explanation
here x=xocos(ωt-π/4) ∴ velocity v=dx/dt=-xoωsin(ωt - π/4) acceleration a=dv/dt Acceleration a=A cos(ωt+δ) comparing the two equations, we get A=xoω2 and δ=3π/4Answer: (a)
Q.23
if x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then which of the following does not change with time ? [ AIEEE 2009]
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a) aT/x
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b) aT+2πv
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c)aT/v
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d)a2T2+4π2v2
Explanation
For an SHM, the acceleration a=-ω2x, where ω2 is a constant. Therefore a/x is constant. The time period T is also constant. Therefore aT/x is constantAnswer: (a)
Q.24
The displacement y of a particle executing periodic motion is given by y=4cos2(t/2)sin(1000t)This expression may be considered to be a result of the superposition of.... independent harmonic motion [ IIT 1992]
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a) two
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b) three
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c)four
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d)five
Explanation
y=4cos2(t/2)sin(1000t)y=2[ 1+cost]sin(1000t)y=2sin1000t + 2costsin1000ty=2sin1000t + [sin(1001t)+cos(999t)]Thus, the given wave comprises of three component waves Answer:(b)
Q.25
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal when displacement ( amplitude=a) is ... [ MPPMT 1987]
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a) a/2
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b) a√2
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c) a/√2
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d) a√2 / 3
Explanation
potential energy=½ mω2y2 Total energy=½ mω2a2 when P.E=K.E then P.E=½ T.E (½) mω2y2=(½)(½) mω2A2 y2=a2/2 y=a/√ 2 Answer: (c)
Q.26
The amplitude of damped oscillator becomes half in one minute. the amplitude after 3 minutes will be 1/X times the original. Where X is [ CPMT 1989]
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a) 2×3
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b) 23
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c) 32
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d) 3×22
Explanation
in a damped oscillations the amplitude decays exponentially . In 1 min amplitude=a/2 In 2 min amplitude=(a/2) ×(1/2)=a/4 In 3 min amplitude=(a/4) ×(1/2)=a/8 ∴ x=23 Answer: (b)
Q.27
A heavy brass-sphere is hung from a spiral spring and it executes vertical vibrations with period T. The ball is now immersed in non-viscous liquid with a density one-tenth that of brass. When set into vertical vibrations with the sphere remaining inside the liquid all the time, the period will be
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a) (9/10) T
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b) T√(10/9)
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c)unchanged
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d)T√(9/10)
Explanation
Periodic time T=2π√(m/k)It is independent of of forceAnswer: (c)
Q.28
A uniform cylinder of length and mass M having cross-sectional area A is suspended with its vertical length, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density d at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is [ IIT 1990]
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a)
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b)
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c)
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d)
Explanation
Let the extension in the spring be 'y'. Then restoring forces on the spring are i) kx due to elastic properties of springii) upthrust=Adgy=weight of liquid displaced ∴ Total restoring force=(ky+Adgy) ∴ M×( acceleration)=- (k+Adg)y acceleration=- (k+Adg)y /M comparing with acceleration=-ω2y we get Answer:(c)
Q.29
A spring has a certain mass suspended from it and its period for vertical oscillations is Tthe spring is now cut into two equal halves and the same mass is suspended from one of the halves. the period of vertical oscillations is now TThe ratio T2 /T1 is ..[ MPPET 1995]
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a) 1/2
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b) 1 /√2
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c) √2
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d) 2
Explanation
Spring constant of the new spring( half part) becomes 2k, where k is spring constant of original spring Answer: (b)
Q.30
The motion of a particle executing simple harmonic motion is given by x=0.01sin[100π(t+0.005)] where x is in meters and t is in seconds. The time period in seconds is [ CPMT 1990]
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a)0.01
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b) 0.02
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c)0.1
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d)0.2
Explanation
Comparing given equation with standard equation x=a sin(ωt+x) we get ω=100π T=2π/ω=2π / 100π=0.02 secAnswer: (b)
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