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Quiz 2
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Q.1
A linear harmonic oscillator of force constant 2×106 N/m and amplitude 0.01m has a total mechanical energy of 160J. Its.. [ CBSE-PMT 1996]
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a) maximum potential energy is 160J
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b) maximum kinetic energy is 100J
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c)minimum potential energy is zero
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d)minimum potential energy is 100J
Explanation
As we know that mechanical energy is equal to total maximum potential energy=160JandMax K.E=½ (ka2)Max K.E.=½ (2×106)×(0.01) 2=100 J Answer:(a,b)
Q.2
A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will [ CBSE-PMT 1997]
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a) first increase and then decrease
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b) first decrease and then increase
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c) go on increasing
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d) go on decreasing
Explanation
Time period of simple pendulum T ∝ √l , here l is the length from centre of gravity of bob Initially, centre of gravity is at the centre of sphere. When water leaks the centre of gravity goes down until it is half filled, Then it begins to go up and finally again goes at the centre So periodic time increases first and then decreases Answer: (a)
Q.3
A body is executing S.H.M When the displacement from the mean position are 4cm and 5cm, the corresponding velocities of the body are 10cm/s and 8cm/s. Then the time period of the body is [ CBSE-PMT 1991]
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a)2π sec
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b) π/2 sec
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c)π sec
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d)(3π /2) sec
Explanation
For SHM equation for velocity at displacement x is given by NowPutting this value in first equationAnswer: (c)
Q.4
The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of π results in the displacement of the particle along.. [ CBSE-PMT 1990]
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a) circle
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b) figures of eight
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c)straight line
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d)ellipse
Explanation
x=asinωt and y=b sin (ωt + π)=-bsinωtThus x/y=-a/b y=(-b/a) x is the equation of straight line Answer: (c)
Q.5
A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from x=A to x=A/2 is [ CBSE PMT -1992]
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a) T/6
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b) T/4
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c)T/3
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d)T/2
Explanation
For S.H.M Time taken to go from x=0 to x=A/2 can be calculated as Total time taken to move from x=0 to x=A is T/4 thus time taken to move from x=A/2 to x=A is t'=T/4 - T/12t'=T/6 and same time will be required to move from x=A to x=A/2 Answer:(a)
Q.6
The potential energy of a long spring when stretched by 2cm is U. If the spring is stretched by 8cm, the potential energy stored in it is .. [ CBSE-PMT 2006]
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a) 8U
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b) 16U
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c) U/4
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d) 4U
Explanation
Potential energy U∝ x2 Since new length is 4 times of initial stretch, thus New Potential energy U'=42U U'=16U Answer: (b)
Q.7
A particle, with restoring force proportional to displacement and resistive force proportional to velocity is subjected to a force FsinωIf the amplitude of the particle is maximum for ω=ω1 and the energy of the particle is maximum for ω=ω2, then [ CBSE-PMT 1989]
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a)ω1=ω0 and ω2 ≠ω0
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b) ω1=ω0 and ω2=ω0
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c)ω1 ≠ ω0 and ω2=ω0
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d)ω1 ≠ ω0 and ω2 ≠ ω0
Explanation
At maximum energy of the particle, velocity resonance takes place, which occurs when frequency of external periodic force is equal to natural frequency of undamped vibrations.i.e ω2=ω0Further, amplitude resonance takes place at a frequency of external force which is less than the frequency of undamped natural vibrations. i.e ω1 ≠ ω0Answer: (c)
Q.8
A particle is executing a simple harmonic motion of amplitude A. Its potential energy is maximum when the displacement from the position of the maximum kinetic energy is .. [ CBSE-PMT 2002]
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a) 0
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b) ±a
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c)±a/2
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d)- a/2
Explanation
Potential energy of particle executing S.H.M=½ (m ω2 x2At x=±A, P.E is max Answer: (b)
Q.9
The displacement of a particle along the x-axis is given by x=a sin2ωt. The motion of the particle corresponds to... [ CBSE-PMT 2010]
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a) simple harmonic motion of frequency ω / π
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b) simple harmonic motion of frequency 3ω / 2π
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c)non simple harmonic motion
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d)simple harmonic motion of frequency ω / 2π
Explanation
If second derivative of x is again in terms of x then function will be S.H.Mx=a sin2ωt Thus equation represents S.H.M and frequency=ω / π Answer:(a)
Q.10
The time period of a simple pendulum is 2 seconds. If its length is increased by 4-times then its period becomes... [ CBSE-PMT 1999]
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a) 16 s
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b) 12 s
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c) 8 s
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d) 4 s
Explanation
periodic time of simple pendulum T ∝ √ l If l is increased by four times period will increase by two times Answer: (d)
Q.11
Two simple pendulums of length 5m and 20m 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.. oscillations[ CBSE-PMT 1998]
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a)5
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b) 1
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c)2
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d)3
Explanation
let the pendulums be in phase after t sec of start. Within this time, if the bigger pendulum executes n oscillations, the smaller one will have executed (n+1) oscillations.Now, the time of n oscillation= Time of (n+1) oscillation of other pendulum=To be in phaseHence number of oscillations executed by shorter pendulum=n+1=1+1=2Answer: (c)
Q.12
A wave has S.H.M whose period is 4 seconds while another wave which also possess SHM has its period 3 seconds. If both are combined, then the resultant wave will have the period equal to ? [ CBSE-PMT 1993]
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a) 4 Sec
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b) 5 sec
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c)12 sec
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d)3 sec
Explanation
Frequency of combined S.H.M Answer: (c)
Q.13
A particle starts simple harmonic motion from mean position. Its amplitude is A and time period is T. What is its displacement when speed is half of its maximum speed [ CBSE-PMT -1996]
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a)
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b)
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c)
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d)
Explanation
vmax=Aω ; When v=vmax/2=Aω/2Now v=ω (A2 -y2)1/2 Answer:(b)
Q.14
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is [ CBSE-PMT 2003] Eis the total energy
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a) E/2
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b) 2E/3
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c) E/8
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d) E/4
Explanation
P.E=E=∝ x2 Thus at half way (x/2), P.E=E/4 Answer: (d)
Q.15
The amplitude of pendulum executing damped simple harmonic oscillation falls to 1/3 of the original value after 100 oscillations. The amplitude fall to S times the original value after 200 oscillations, where S is ...[ CBSE-PMT 2002]
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a)1/9
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b) 1/2
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c)2/3
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d)1/6
Explanation
In damped harmonic oscillator, amplitude falls exponentiallyAfter 100 oscillations amplitude falls to 1/3 times∴ After next 100 oscillation, that is after 200 oscillations amplitude falls to (1/3) 2=1/9 timesAnswer: (a)
Q.16
A mass m is vertically suspended from a spring of negligible mass, the system oscillates with frequency n. What will be the frequency of system, if a mass 4m is suspended from same spring? [ CBSE-PMT 1998]
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a) n/4
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b) 4n
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c)n/2
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d)2n
Explanation
We know that frequency n ∝ 1/√m Thus if mass is 4 m then new frequency will be half of earlier=n/2Answer: (c)
Q.17
A particle is subject to two mutually perpendicular simple harmonic motions such that its x and y coordinates are given byx=2sinωt and y=2sin(ωt + π/4) The path of the will be .. [CBSE-PMT 1994]
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a)a straight line
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b) a circle
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c)an ellipse
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d)a parabola
Explanation
As a phase difference is π/4, the resultant path of particle is an ellipse Answer:(c)
Q.18
A particle executing S.H.M has amplitude 0.01m and frequency 60 Hz. The maximum acceleration of the particle is ... [ CBSE-PMT 1999]
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a) 144 π2 m/s2
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b) 120 π2 m/s2
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c) 80 π2 m/s2
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d) 60 π2 m/s2
Explanation
Amplitude (A)=0.01m, frequency=60 Hz Maximum Acceleration=Aω22Maximum Acceleration=0.01 × (2πn)2 =144π2 m/sec2 Answer: (a)
Q.19
In case of forced vibration, the resonance wave becomes very sharp when the .. [ CBSE-PMT 2003]
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a)quality factor is small
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b) damping force is small
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c)restoring force is small
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d)applied periodic force is small
Explanation
The resonance curve becomes very sharp when damping force is smallAnswer: (b)
Q.20
Two simple harmonic motions acts on a particle. These harmonic motions are x=A cos(ωt + δ), y=A cos(ωt + α) when δ=α + π/2, the resulting motion is .. [ CBSE-PMT 2000]
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a) a circle and the actual motion is clockwise
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b) an ellipse and the actual motion is counterclockwise
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c)an ellipse and the actual motion is clockwise
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d)a circle and the actual motion is counter clockwise
Explanation
x=A cos(ωt + δ) When δ=α + π / 2x=-A sin ( ωt + α) -- eq(1) y=A cos (ωt + α) --eq(2) Squaring equation 1 and 2 and then adding we getx2 + y2=A2[cos2(ωt + α) + sin2(ωt + α)]x2 + y2=A2Above equation represents circle. The present motion is anticlock wise Answer: (d)
Q.21
The particle executing simple harmonic motion has a kinetic energy K0cos2ωt. The maximum value of the potential energy and the total energy are respectively.. [ CBSE-PMT 2007]
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a) K0 / 2 and K0
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b)K0 and 2K0
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c)K0 and K0
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d)0 and 2K0
Explanation
When ωt=0 cos2ωt=1Thus Maximum Kinetic energy=K0Now Δ K.E=ΔP.E∴ maximum potential energy=K0 Answer:(c)
Q.22
A simple pendulum performs simple harmonic motion about x=0 with an amplitude 'a' and time period T. The speed of the pendulum at x=a/2 will be.. [ CBSE-PMT 2009]
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a)
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b)
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c)
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d)
Explanation
Speed When x=a/2 Answer: (c)
Q.23
A particle is executing S.H.M of amplitude a and time period=4 sec. then the time taken by it to move from the extreme position to half of the amplitude is [ BHU 1995]
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a)1 sec
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b) 1/3 sec
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c)2/3 sec
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d)4/3 sec
Explanation
At extreme position y=a y=a sinωt1 or a=a sinωt1 or ωt1=π/2 or t1=π/2ωAt half amplitude a /2=a sinωt21/2=sinωt2 ωt2=π/6 or t2=π/6ω ∴ Δt=π/2ω - π/6ω=π/3ω But ω=2π / T Δt=(π/3)(T/ 2π)=T/6 Δt=4/6=2/3 second Answer: (c)
Q.24
Two bodies M and N of equal masses are suspended from two separate massless springs of spring constant k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of M to that of N is [ IIT 1988]
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a)
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b)
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c)
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d)
Explanation
formula for velocity v=aω=(a×2π) / TBoth the bodies have same velocity thus Periodic time T ∝ 1 /√k thus Answer: (d)
Q.25
A particle free to move along the x-axis has potential energy given by U(x)=k[-exp(-x2)] for -∞ ≤x≤+∞, where k is a positive constant of appropriate dimensions. Then [ IIT 1999]
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a) at points away from the origin, the particle is in stable equilibrium
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b) for any finite nonzero value of x, there is a force directed away from the origin
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c)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.
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d)for small displacement from x=0, the motion is simple harmonic
Explanation
By taking derivative of U with respect to x we get From the above equation force is proportional to displacement and direction of force is toward equilibrium. Answer: (d)
Q.26
The period of oscillation of simple pendulum of length L suspended from the roof of a vehicle which moves without friction on an inclined plane of inclination α, is given by [ IIT 2000]
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a)
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b)
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c)
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d)
Explanation
Pendulum will experience pseudo acceleration of gsinα in upward direction parallel to inclined plane as shown in figure. Component of pseudo acceleration along x axis will ne gsinαcosα and component along y axis will be gsin2α Thus acceleration along y axis=g-gsin2α=gcos2α resultant acceleration a Answer:(a)
Q.27
A simple pendulum has time period Tthe point of suspension is now moved upward according to the relation y=Kt2 (K=1 m/s2) where y is the vertical displacement. The time period now becomes T2 . The ratio of T12 / T22 is ... ( g=10 m/s2)
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a) 5/6
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b) 6/5
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c) 1
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d) 4/5
Explanation
given displacement y=kt2 ∴ dy/dt=2kt or d2y / dt2=2k k=1 given a=2 m/s2 Now effective acceleration g'=g+a=12 m/s2 We know that T=2π √(l/g) ∴ T12 / T22=g' / g=12/10=6/5 Answer: (b)
Q.28
The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is ..[ IIT 2009]
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a)
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b)
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c)
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d)
Explanation
If the spring constant k1 is compressed by x1 and the spring constant k2 is compressed by x2 If total compression is A then x1 + x2=A --eq(1) since same force acts on both the springs k1x1=k1x1Thus x2=k1x1 / k2 --eq(2)from equation 1 and 2 we get x1 + k1x1 / k2=Ax1=k2A / (k2+k1)Answer: (d)
Q.29
A uniform rod of length and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constant k. The springs are fixed to rigid supports as shown in figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is [ IIT 2009]
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a)
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b)
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c)
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d)
Explanation
As shown in figure both the springs are stretched by lθ/2Restoring torque=-(Restoring force )×perpendicular distance Restoring torque=- k(lθ)/2 × (l/2)=-(kl2θ/ 4)Therefore total restoring torque due to both the spring=-2(kl2θ/ 4)Restoring torque=-(kl2θ/ 2)If I is moment of inertia then τ=Id2θ / dt2 Thus comparing above equation with standard equation for SHM we get Answer: (c)
Q.30
In a simple harmonic oscillator, at mean position [ AIEEE 2002]
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a) kinetic energy is minimum, potential energy is maximum
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b) both kinetic and potential energies are maximum
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c)kinetic energy is maximum, potential energy is minimum
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d)both kinetic and potential energies are minimum
Explanation
Answer:(c)
0 h : 0 m : 1 s
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