71 + Interview Questions in OSCILLATION-AND-SIMPLE-HARMONIC-MOTION IN GENERAL AWARENESS Page 2 InterviewSolution

51.	A boby is moving in a room with a velocity of `20m//s` perpendicular to the two walls separated by 5 meters .There is no friction and the collisionnn with the walls are elastic.A. Not periodicB. Periodic but not simple harmonicC. Priodic and simple harmonicD. Periodic with variable time period
Answer» Correct Answer - B Body collides elastically with wolls of room So, will be no loss energy and it will remain colliding with wolts of room, so its motion will be periodic .Then is no change in energy of the body hence there is no acceleration so it motion to not `SHM`

Discussion

52.	Which of the following expression does not repressent SHM?A. `A cos omega t`B. `A sin2 omega t`C. `A sin omega t + B cos omega t `D. `A sin^(2) omega t`
Answer» Correct Answer - D For `SHM (d^(2)y)/(dt^(2)) prop - y` If `y = A sin^(2) omega t` `(dy)/(dt) = (2A sin omega t cos omega t ) omega` ltbnrgt `= A omega sin 2 omega t` `(d^(2)y)/(dt^(2)) = + 2 A omega^(2) cos 2 omega t` `i.e. (d^(2)y)/(dt^(2))` is not proportional to `- y`

Discussion

53.	A particle executes simple harmonic motion according to the displacement equation `y = 10 cos(2 pi t + (pi)/(6))cm` where `t` is in second The velocity of the particle at `t = 1/6` second will beA. `-6.28ms^(-1)`B. `-0.628ms^(-1)`C. `0.628ms^(-1)`D. `6.28ms^(-1)`
Answer» Correct Answer - C Velocity `v = (dv)/(dt) = (d)/(dt) [10 cos(2 pi t+ (pi)/(6))]` `= - 20pi sin (2pi t + (pi)/(6)) cms^(-1)` `v((1)/(6)) = - 20(3.14)sin((pi)/(3) + (pi)/(6)) cms^(-1)` `= 62.8 cms^(-1)("numirically")[Using sin"(pi)/(2) = 1]` `= 0.628ms^(-1)`

Discussion

54.	A block of mass `2kg` executes simple harmonic motion under the reading from at aspring .The angular and the time period of motion are `0.2 cm` and `2pi`sec respectively Find the maximum force execut by the spring in the block.A. `0.05 N`B. `0.002 N`C. `0.003 N`D. `0.004 N`
Answer» Correct Answer - D The mass of block `= m = 2kg amp = A = 0.2cm` and time period `T = 6.28 sec` The maximum force exerted will be `KA` and it will be at extreme position. `f_(max) = m omega^(2) A [omega = sqrt((K)/(m)) "where" K = "spring constant"]` `= 2 xx ((2pi)/(6.28))^(2) xx (0.2 xx 10^(-2)) = 0.004 N` Acceleration at `0.1 cm` is given by Acceleration `a= - omega^(2)x = - ((2pi)/(6.28))^(2) xx (0.1cm)` ` = -0.1 cm//sec = - 10^(-3) m//s^(2)`

Discussion

55.	The displacement of a perticle varies with time as `x = 12 sin omega t - 16 sin^(2) omega t` (in cm) it is motion is `S.H.M.` then its maximum acceleration isA. `12 omega^(2)`B. `36 omega^(2)`C. `133 omega^(2)`D. `sqrt(192) omega^(2)`
Answer» Correct Answer - B `x = 12 sin omega t - 16 sin^(3) omega t = 4(3 sin omega t - 4 sin ^(3)omega t]` `= 4[sin 3 omega t]` (By using `sin 3 theta = 3 sin theta - 4 sin ^(3) theta`] `:.` Maximum acceleration `A_(max) = (3omega)^(2) xx 4 = 36 omega^(2)`.

Discussion

56.	Two simple harmonic motion of angular frequency `100and 1000 rads^(-1)` have the same displacement amplitude The ratio of their maximum acceleration isA. `1 : 10`B. `1 : 10^(2)`C. `1 : 10^(3)`D. `1 : 10^(4)`
Answer» Correct Answer - B Acceleration of simple harmonic motion is `a_(max) = - omega^(2)A` `or ((a_(max)))/((a_(max)_(1)))=(omega_(1)^(2))/(omega_(2)^(2))` (as A remain the same) or `((a_(max)))/((a_(max)_(1))) = ((100)^(2))/((1000)^(2)) = ((1)/(10))^(2) = 1.10^(2)`

Discussion

57.	The displacement of a particle varies according to the relation `y = 4(cos pi t + sin pi t)`. The amplitude of the particle isA. 8B. -4C. 4D. `4sqrt(2)`
Answer» Correct Answer - D For given relation Resultant amplitude `= sqrt(4^(2) +4^(2)) + 4 sqrt(2)`

Discussion

58.	Assertion: In a simpleharmonic motion the kinetic and potential energy becomes equal when the displacement is `(1)/(sqrt(2))` time the amplitude Reason: is `SHM` kinetic energy is zero when potential energy is maximumA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false
Answer» Correct Answer - B From the relation potential energy of `SHM` is given `as PE = (1)/(2) m omega^(2)y^(2)` `andKE = (1)/(2)m omega^(2)(A^(2) - y^(2))` Where `y` is the displacement and `A` is the amplitude by the condition `KE = PE` `(1)/(2)m omega^(2)(A^(2) - y^(2)) = (1)/(2) m omega^(2)y^(2)` `rArr m omega^(2)y^(2)A^(2)` `rArr y = (1)/(sqrt(2))momega^(2)A^(2)` Also we know that total energy is `K = KE + PE` Hence in `PE` is maximum then `KE` will be zero

Discussion

59.	A particle of mass `m` oscillates with simple harmonic motion between points `x_(1)` and `x_(2)` the equilibrium position being `O` its potential energy in plotted it will be as given bellow in the graphA. B. C. D.
Answer» Correct Answer - D Potential energy of particle performing `SHM` is given by `PE = (1)/(2) m omega^(2)y^(2)` i.e. veries parabolicalty such thet at mean position is become zero maximum at exterme position

Discussion

60.	when two displacements represented by `y_(1) = a sin(omega t)` and `y_(2) = b cos (omega t)` are superimposed the motion isA. Not a simple harmonicB. simple harmonic with amplitude`(a)/(b)`C. simple harmonic with amplitude`sqrt((a)^(2) +(b)^(2))`D. simple harmonic with amplitude`((a+b))/(2)``
Answer» Correct Answer - C The two displacement equation are `y_(1) = a sin(omega t)` ltbrtgt and `y_(2) = b cos(omega t) = b(omega t + (pi)/(2))` `y_(eq) = y_(1) + y_(2)` `= a sin omega t + b cos omega t` `= a sin omega t + b sin(omega t + (pi)/(2))` Since the frequency for both `SHM` are same resultant motion will be `SHM` Now `A_(eq) = sqrt(a^(2) +b^(2) + 2ab cos (pi)/(2))` `rArr sqrt(a^(2) +b^(2))`

Discussion

61.	The `KE` and `PE` , at is a particle executing `SHM` amplitude `A` will be equal when its displacement isA. `Asqrt(2)`B. `A//2`C. `A//sqrt(2)`D. `Asqrt(2//3)`
Answer» Correct Answer - C `K = (1)/(2) m omega ^(2)(A^(2)- y^(2))` `U = (1)/(2) m omega^(2)y^(2)` `K = U` or `(1)/(2) m omega^(2) (A^(2) - y^(2)) = (1)/(2) m omega^(2) y^(2)` `i.e. 2y^(2) = A^(2)` or `y = (A)/(sqrt(2))`

Discussion

62.	Assertion : The periodic time of hard spring is less its compared to that of soft spring Reason: The periodic time depend upon the spring constantA. If both assertion and reason are true and the reasopn is a true explanation of the assertionB. If both assertion and reason are true and but the reason is not the correct explanation of assertionC. If the assertion is true but reason is falseD. If both the assertion and reason are false
Answer» Correct Answer - A The time period of a oscillation spring is given by `T = 2pi sqrt((m)/(k)) rArr T prop (1)/(sqrt(k)) `since the spring constant is large for hard spring therefore hard spring has a less periodic time as compared to soft spring

Discussion

63.	Assertion : The periodic time of hard spring is less its compared to that of soft string Reason: The periodic time depend upon the spring constantA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false
Answer» Correct Answer - B From the relation `T = 2pi sqrt((m)/(k))`, where k is the spring constant `T = (1)/(sqrt(k))`…(1) As we know that the spring constant of heat spring large in comparing in that of the spring therefore from Eq. (1) spring is large in comparison to that at soft spring therefore from Eq. (1) the time period will be less then for hard spring in comparision to that of soft spring

Discussion

64.	A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and energy (T.E) are measured as a function of displacement `x`. Which of the following staements is true?A. `PE` is maximum when `x= 0`B. `KE` is maximum when `x= 0`C. `TE` is zero when `x= 0`D. `KE` is maximum when `x` is maximum
Answer» Correct Answer - B in `S.H.M` at mean position i.e. at `x= 0` kinetic energy will be maximum and `pE` will be minimum Total energy is always constant

Discussion

65.	Assertion : Damped vibrations indicate loss of energy Reason : The loss may be due to friction , air resistance ectA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false
Answer» Correct Answer - A In nature many types of direspating force such as friction air resistance etc., are operating Therefore As the oscilates, a part of its energy is used up in overcoming these desipoting force. such oscillation are damped oscillations.

Discussion

66.	The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will beA. `2sqrt(3) cm`B. `sqrt(3) cm`C. `1 cm`D. `2 cm`
Answer» Correct Answer - D At mean position velocity is maximum `i.e. v_(max) = omega a rArr omega rArr omega = (v_(max))/(a) = (16)/(4) = 4` `:. V = omega sqrt(a^(2) - y^(2)) rArr 8 sqrt(4^(2) - y^(2))` `rArr 192 = 16(16 - y^(2)) rArr 12 = 16 - y^(2) rArr y = 2 cm`

Discussion

67.	Assertion : In `SHM` acceleration is always direction toward the mean position Reason : The body stope momentally at the extrame position and then moves back to mean positionA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false
Answer» Correct Answer - A In `SHM` the bob (body) does a to fro motion about a fixed point called the mean position At the extreme position velocity of bob is zero but acceleration is maximum which lends it the to moven back to mean position .Thus acceleration is always direction toward mean position `a prop - x` Which state that acceleration is in oppasite direction of displacement i.e. toward mean position

Discussion

68.	In the figure shown a block of masss `m` is atteched at ends of two spring The other ends of the spring are fixed The mass `m` is released in the vertical plane when the spring are released The velocity of the block is maximum when A. `k_(1)` is compressed and `k_(2)` is elongatedB. `k_(1)` is elongated and `k_(2)` is compressedC. `k_(1)` and `k_(2)` both are compressedD. `k_(1)` and `k_(2)` both are elongated.
Answer» Correct Answer - B Speed of block is maximum at mean position. As mean position upper spring is extended and lower spring is compressed.

Discussion

69.	Two very small having mass `m` are atteched to two masses rods of length `l` Now these rods are joined it form `V` like figure having angle `60^(@)` This assumility is new higest in a verticle plane so that it can rotan without any friction about a horizontal axis perpendicular in the plane of figure) as shown in the figure The period of small oscilation of this asseamble is A. `2pi sqrt((2l)/(g))`B. `2pi sqrt((2l)/(sqrt(3)g))`C. `2pi sqrt((1)/(sqrt(3)g))`D. `2pisqrt((sqrt(3)l)/(g))`
Answer» Correct Answer - B `T = 2pi sqrt((2ml^(2))/(2mgl))cos 30^(@) = 2pi sqrt((2l)/(gsqrt(3))`

Discussion

70.	What will be the period of the displacement body of mass `m`? A. `2pisqrt((m)/(2K))`B. `2pisqrt((3m)/(K))`C. `2pisqrt((3m)/(2K))`D. `pisqrt((3m)/(K))`
Answer» Correct Answer - C Equivalent force constant of three spring `k_(eq) = (k_(1)k_(2))/(k_(1) + k_(2)) = (k xx 2k)/(3k) = (2k)/(3)` `T = 2pi sqrt((m)/(k_(eq))) = 2pi sqrt((m)/(2k//3))` `rArr T = 2pi sqrt((3m)/(2k))`

Discussion

71.	What will be the force constant of the spring system shown in figure? A. `[(1)/(k_(1)) +(1)/(k_(2))]`B. `[(1)/(2k_(1)) +(1)/(k_(2))]^(-1)`C. `[(1)/(k_(1)) +(1)/(k_(2))]^(-1)`D. `[(1)/(2k_(1)) +(1)/(k_(2))]`
Answer» Correct Answer - B For parrallel combinacation of first two identical spring of spring constant `k_(1)` effactive spring constant `K_(p) = 2k_(1)` Now spring of spring constant `k_(p)` and `k_(2)` are joined in spring so the force constant of the spring constant of the system is `(1)/(k_(s)) = (1)/(k_(p)) + (1)/(k_(2)) = ((1)/(2k_(1)) + (1)/(k_(2)))^(-1)`

Discussion

Explore topic-wise InterviewSolutions in .

A boby is moving in a room with a velocity of `20m//s` perpendicular to the two walls separated by 5 meters .There is no friction and the collisionnn with the walls are elastic.A. Not periodicB. Periodic but not simple harmonicC. Priodic and simple harmonicD. Periodic with variable time period

Which of the following expression does not repressent SHM?A. `A cos omega t`B. `A sin2 omega t`C. `A sin omega t + B cos omega t `D. `A sin^(2) omega t`

A particle executes simple harmonic motion according to the displacement equation `y = 10 cos(2 pi t + (pi)/(6))cm` where `t` is in second The velocity of the particle at `t = 1/6` second will beA. `-6.28ms^(-1)`B. `-0.628ms^(-1)`C. `0.628ms^(-1)`D. `6.28ms^(-1)`

A block of mass `2kg` executes simple harmonic motion under the reading from at aspring .The angular and the time period of motion are `0.2 cm` and `2pi`sec respectively Find the maximum force execut by the spring in the block.A. `0.05 N`B. `0.002 N`C. `0.003 N`D. `0.004 N`

The displacement of a perticle varies with time as `x = 12 sin omega t - 16 sin^(2) omega t` (in cm) it is motion is `S.H.M.` then its maximum acceleration isA. `12 omega^(2)`B. `36 omega^(2)`C. `133 omega^(2)`D. `sqrt(192) omega^(2)`

Two simple harmonic motion of angular frequency `100and 1000 rads^(-1)` have the same displacement amplitude The ratio of their maximum acceleration isA. `1 : 10`B. `1 : 10^(2)`C. `1 : 10^(3)`D. `1 : 10^(4)`

The displacement of a particle varies according to the relation `y = 4(cos pi t + sin pi t)`. The amplitude of the particle isA. 8B. -4C. 4D. `4sqrt(2)`

A particle of mass `m` oscillates with simple harmonic motion between points `x_(1)` and `x_(2)` the equilibrium position being `O` its potential energy in plotted it will be as given bellow in the graphA. B. C. D.

when two displacements represented by `y_(1) = a sin(omega t)` and `y_(2) = b cos (omega t)` are superimposed the motion isA. Not a simple harmonicB. simple harmonic with amplitude`(a)/(b)`C. simple harmonic with amplitude`sqrt((a)^(2) +(b)^(2))`D. simple harmonic with amplitude`((a+b))/(2)``

The `KE` and `PE` , at is a particle executing `SHM` amplitude `A` will be equal when its displacement isA. `Asqrt(2)`B. `A//2`C. `A//sqrt(2)`D. `Asqrt(2//3)`

The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will beA. `2sqrt(3) cm`B. `sqrt(3) cm`C. `1 cm`D. `2 cm`

What will be the period of the displacement body of mass `m`? A. `2pisqrt((m)/(2K))`B. `2pisqrt((3m)/(K))`C. `2pisqrt((3m)/(2K))`D. `pisqrt((3m)/(K))`

What will be the force constant of the spring system shown in figure? A. `[(1)/(k_(1)) +(1)/(k_(2))]`B. `[(1)/(2k_(1)) +(1)/(k_(2))]^(-1)`C. `[(1)/(k_(1)) +(1)/(k_(2))]^(-1)`D. `[(1)/(2k_(1)) +(1)/(k_(2))]`