956 + Interview Questions in OSCILLATIONS IN GENERAL AWARENESS Page 17 InterviewSolution

801.	A particle executing SHM has a frequency of 10 Hz when it crosses its equilibrium position with a velocity of `2pi m//s`. Then the amplitude of vibration isA. 0.1 mB. 0.2 mC. 0.4 mD. 1 m
Answer» Correct Answer - A `v_(m)=Aomega` `therefore A=(v_(m))/(2pin)=(2pi)/(2pixx10)=0.1m`

Discussion

802.	A simple harmonic motion is represented by x(t) = 10 sin(20t + 0.5)Find its amplitude, frequency and initial phase.
Answer» Comparing the given equation with standard equation of S.H.M. x(t) = a sin(ωt + ϕ_o) (i) Amplitude, a = 10 m (ii) Angular frequency, ω = 20 rad s^-1 (iii) Frequency is given by v = \(\frac{ω}{2\pi}\) = 3.18 Hz (iv) Initial phase, ϕ_o = 0.5 rad.

802.

A simple harmonic motion is represented by x(t) = 10 sin(20t + 0.5)Find its amplitude, frequency and initial phase.

Answer»

Comparing the given equation with standard equation of S.H.M.

x(t) = a sin(ωt + ϕ_o)

(i) Amplitude, a = 10 m

(ii) Angular frequency,

ω = 20 rad s^-1

(iii) Frequency is given by

v = \(\frac{ω}{2\pi}\) = 3.18 Hz

(iv) Initial phase, ϕ_o = 0.5 rad.

Discussion

803.	A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre isA. `sqrt(2) pi`B. `16 sqrt(2) pi`C. `24 sqrt(2) pi`D. `32 sqrt(2)//pi`
Answer» Correct Answer - D For simple harmonic motion, y = a sin `omega`t `therefore" "y = a sin ((2pi)/(T))t" "("at t"=2s)` `y_(1)=a sin [((2pi)/(16))xx2]=a sin((pi)/(4))=(a)/(sqrt(2))" "...(i)` At t = 4 s or after 2 s from mean position. `y_(1)=(a)/(sqrt(2))"and velocity"=4 ms^(-1)` `therefore` Velocity `= omega sqrt(a^(2)-y_(1)^(2))` or `4 = ((2pi)/(16))sqrt(a^(2)-(a^(2))/(2))" "["From Eq. (i)"]` or `4 = (pi)/(8) xx (a)/(sqrt(2))or a = (32 sqrt(2))/(pi)m`

Discussion

804.	Time period of a particle executing SHM is 16s.At time `t = 2s`, it crosses the mean position . Its amplitude of motion is `( 32sqrt(2))/(pi) m`. Its velocity at `t = 4s` isA. `1 ms^(-1)`B. ` 2ms^(-1)`C. ` 4ms^(-1)`D. `8 ms^(-1)`
Answer» Correct Answer - 3

Discussion

805.	The equation of motion of a particle started at `t=0` is given by `x=5sin(20t+pi//3)` where x is in centimeter and t in second. When does particle (a) first come to rest (b) first have zero acceleration (c) first have maximum speed?A. `(pi)/(90)sec`B. `(pi)/(120)sec`C. `(pi)/(60)sec`D. `(pi)/(30)sec`
Answer» Correct Answer - D `a = (d^(2)y)/(dt^(2)) =0` at mean position, particle is initially moving towards positive exterme.

Discussion

806.	The potential energy of a particle oscillating on x-axis is given as `U +20 +(x-2)^(2)`. The mean position is atA. `x =2m`B. `x =1m`C. `x =3m`D. `x =4m`
Answer» Correct Answer - A `U` is minimum at mean position so mean position is at `x = 2m`

Discussion

807.	A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre isA. `sqrt(2)pi`B. `16sqrt(2)pi`C. `32sqrt(2)//pi`D. `4//pi`
Answer» Correct Answer - C `y_(1) = A sin ((2pi)/(T)xxt), V = omega sqrt(A^(2) - y_(1)^(2))`

Discussion

808.	A simple pendulum of length `l` is connected to the ceiling of a vehicle that is moving down along a smooth inclined plane `4` in `5`. Then its period of oscillation isA. `2pi sqrt((5l)/(4g))`B. `2pisqrt((4l)/(5g))`C. `2pisqrt((5l)/(3g))`D. `2pi sqrt((3l)/(5g))`
Answer» Correct Answer - C `T = 2pi sqrt((l)/(g cos theta))`

Discussion

809.	A particle starts SHM from mean position O executing SHM A and B are the two point at which its velocity is zero . It passes through a certain point P at time `t_(1) = 0.5 and t_(2) = 1.5 s` with a speed of `3 m//s`. i. The maximum speed ....... ii. ratio `AP//PB` .....A. `3m//s`B. `3sqrt(2)m//s`C. `3sqrt(3)m//s`D. `6m//s`
Answer» Correct Answer - B `T = 2sec, v = omega sqrt(A^(2) -x^(2)), v = 3m//sec` at `x = (A)/(sqrt(2)), v_(max) = A omega`

Discussion

810.	The relation between acceleration and displacement of four particles are given below. The particle undergoing SHM is:A. `a_(x) = +2x`B. `a_(x) =+2x^(2)`C. `a_(x) =- 2x^(2)`D. `a_(x)= -2x`
Answer» Correct Answer - D In option `(d)a_(x) = - 2x` or `a alpha -x`, the acceleration of the particle is proportional to negative of displaement. Hence it represnts `S.H.M`.

Discussion

811.	A block is kept on a rough horizontal plank. The coefficient of friction between block and plank is `1//2`. Plank is undergoing `SHM` is angular frequency `10 rad//s`. Find the maximum amplitude of plank in which the block does slip over plank `(g = 10 m//s^(2))`.
Answer» Maximum acceleration in `SHM` is `a_(max) = •^(2)A` this will be provided to the block by friction. Hence, `a_(max) = •g or •^(2)A = •g` or `A = (•g)/(•^(2)) = (((1)/(2))(10))/((10)^(2)) =0.05m = 5cm`

Discussion

812.	Give some practical examples of S. H. M?
Answer» Some practical examples of S. H. M. are :- 1) Motion of piston in a gas – filled cylinder. 2) Atoms vibrating in a crystal lattice. 3) Motion of helical spring.

812.

Give some practical examples of S. H. M?

Answer»

Some practical examples of S. H. M. are :-

1) Motion of piston in a gas – filled cylinder.

2) Atoms vibrating in a crystal lattice.

3) Motion of helical spring.

Discussion

813.	In 20 s, two simple pendulums, P and Q, complete 9 and 7 oscillations, respectively, on the Earth. On the Moon, where the acceleration due to gravity is \(\frac16\)th that on the Earth, their periods are in the ratio (A) 8 : 1 (B) 9 : 7 (C) 7 : 9 (D) 3 : 14.
Answer» Correct option is (C) 7 : 9

Discussion

814.	A seconds pendulum is suspended in an elevator moving with a constant speed in the downward direction. The periodic time (T) of that pendulum is (A) less than two seconds (B) equal to two seconds (C) greater than two seconds (D) very much greater than two seconds.
Answer» (B) equal to two seconds

Discussion

815.	If T is the time period of a simple pendulum in an elevator at rest, its time period in a freely falling elevator will be(A) \(\frac{T}{\sqrt2}\)(B) √2T (C) 2T (D) infinite.
Answer» Correct option is (D) infinite.

Discussion

816.	STATEMENT-1`:` Time periodof oscillation of a simple pendulum mounted in a cabin that is freely falling is zero and STATEMENT -2 `,` In the cabin falling freely under gravity the pendulum is in state of weightlessness.
Answer» Correct Answer - 4

Discussion

817.	Time period of a simple pendulum in a freely falling lift will beA. FiniteB. InifiniteC. ZeroD. All of these
Answer» Correct Answer - 2

Discussion

818.	The displacement of a particle making S.H.M. is given by `x=6cos(100t+(pi)/(4))m` then the frequency isA. `1.592 Hz`B. `15.92 Hz`C. `159.2 Hz`D. `1592 Hz`
Answer» Correct Answer - B `x=6cos(100t+(pi)/(4))` `omega=100` `2pin=100` `n=(100)/(2pi)=(50)/(pi)=15.92Hz`.

Discussion

819.	A particle oscillating under a force `vecF=-kvecx-bvecv` is a (k and b are constants)A. Linear oscillationB. Forces oscillationsC. Damped oscilationD. SHM
Answer» Correct Answer - 3

Discussion

820.	The displacement of a particle of mass 100 g executing SHm is given by the equation y = 0.2 cos `(100t+(pi)/(4))` meter. Its total energy isA. 1 JB. 10 JC. 2 JD. 20 J
Answer» Correct Answer - D `E=(1)/(2)momega^(2)A^(2)=(1)/(2)xx0.1xx100^(2)xx(0.2)^(2)` `=20J`

Discussion

821.	Displacement of a particle executing SHM s `x= 10 ( cos pi t + sin pi t)`. Its maximum speed isA. `5 pi m//s`B. ` 10 pi m//s`C. ` 10 sqrt(2) pi m//s`D. ` 5 sqrt(2 ) pi m//s`
Answer» Correct Answer - 3

Discussion

822.	Displacement time equation of a particle executing `SHM` is, `x = 10 sin ((pi)/(3)t+(pi)/(6))cm`. The distance covered by particle in `3s` isA. `5cm`B. `20cm`C. `10cm`D. `15cm`
Answer» Correct Answer - B `X = A sin (omegat +ph), T = (2pi)/(omega)`

Discussion

823.	Displacement of a particle executing SHM s `x= 10 ( cos pi t + sin pi t)`. Its maximum speed isA. 10 cmB. 20 cmC. `5 sqrt(2)` cmD. 50 cm
Answer» Correct Answer - A Here, `x = 5 sqrt(2) (sin 2 pi t + cos 2 pi t)` `rArr" "x = 5 sqrt(2) sin 2 pi t + 5 sqrt(2) cos 2 pi t" "...(i)` The standard equation of simple harmonic motion is given by `x = A_(1) sin omega t + A_(2) cos omega t" "...(ii)` Now, comparing Eqs. (i) and (ii), we obtain `A_(1)=5 sqrt(2) and A_(2)=5 sqrt(2)` So, the resultant amplitude of the motion `A = sqrt(A_(1)^(2)+A_(2)^(2))=sqrt((5 sqrt(2))^(2)+(5 sqrt(2))^(2))= 10` cm

Discussion

824.	Time period of a simple pendulum will be double, if weA. decrease the length 2 timesB. decrease the length 4 timesC. increase the length 2 timesD. increase the length 4 times
Answer» Correct Answer - D The formula for time period is, `T = 2pi sqrt((l)/(g)) prop sqrt(l)` If the length is increased by four times, then time period will be doubled.

Discussion

825.	The velocity of a particle performing simple harmonic motion, when it passes through its mean position iA. infinityB. zeroC. minimumD. maximum
Answer» Correct Answer - D The relation for velocity of a particle executing SHM is given by, `v = omega sqrt(A^(2) - y^(2))`. Hence, at mean position y = 0. The velocity is maximum. `V_(max) = A omega` (where, A is amplitude)

Discussion

826.	Two springs, of spring constants `k_(1)` and `K_(2)`, have equal highest velocities, when executing SHM. Then, the ratio of their amplitudes (given their masses are in the ratio `1:2`) will beA. `((k_(2))/(k_(1)))^(1//2)`B. `((k_(1))/(k_(2)))^(1//2)`C. `(k_(1))/(k_(2))`D. `k_(1)k_(2)`
Answer» Correct Answer - A The angular frequency of spring is given by `omega = sqrt((k)/(m)) prop sqrt(k)` For equal maximum velocities, we have `A_(1)omega_(1) = A_(2)omega_(2)` or `(A_(1))/(A_(2))=(omega_(2))/(omega_(1))=sqrt((k_(2))/(k_(1)))=((k_(2))/(k_(1)))^((1)/(2))" "(because m = m_(1) = m_(2))`

Discussion

827.	Two springs of constants `k_1` and `k_2` have equal maximum velocities, when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will beA. `(k_(1))/(k_(2))`B. `(k_(2))/(k_(1))`C. `((k_(1))/(k_(2)))^(1//2)`D. `((k_(2))/(k_(1)))^(1//2)`
Answer» Correct Answer - D `v_(1)=v_(2)` `A_(1)omega_(1)=A_(2)omega_(2)` `(A_(1))/(A_(2))=(omega_(2))/(omega_(1))=(T_(1))/(T_(2))=sqrt((k_(2))/(k_(1)))`

Discussion

828.	A spring of force constant `600 Nm^(-1)` is mounted on a horizontal table. A mass of 1.5 kg is attached to the free end of the spring,pulled sideways to a distance of 2 cm and released . The speed of the mass when the spring is compressed by 1 cm isA. `0.175 ms^(-1)`B. ` 0. 35 ms^(-1)`C. ` 0.7 ms^(-1)`D. ` 1.4 ms^(-1)`
Answer» Correct Answer - 2

Discussion

829.	A spring of force constant `600 Nm^(-1)` is mounted on a horizontal table. A mass of 1.5 kg is attached to the free end of the spring, pulled sideways to a distance of 2 cm and released .P.E. of the mass when it momentarily comes to rest and total energy areA. `0.12 J , 0`B. ` 0,0.12 J`C. `0,0`D. ` 0.12 J , 0.12J`
Answer» Correct Answer - 4

Discussion

830.	If the earth were a homogeneous sphere of radius R and a straight hole bored in it through its centre, show that a body dropped into the hole will execute SHM and find its time period.
Answer» Correct Answer - 5077.6s

Discussion

831.	If a hole is bored along the diameter of the earth and a stone is dropped into holeA. reaches the centre of the earth and stopsB. reaches the other side of the earth and stopsC. executes simple harmonic motion about the centre of the earthD. reaches the other side of the earth and escapes into space
Answer» Correct Answer - C

Discussion

832.	The total energy of a simple pendulum is x. When the displacement is half of amplitude,its KE will beA. `(x)/(2)`B. `(x)/(4)`C. `(3x)/(4)`D. `x`
Answer» Correct Answer - 3

Discussion

833.	Two identical springs have the same force constant of `147Nm^(-1)`. What elongation will be produced in each spring in each case shown in figure? `g=9.8ms^(-2)`. A. `(1)/(6)m,(2)/(3) m, (1)/(3)m`B. `(1)/(3)m,(1)/(3)m,(1)/(3)m`C. `(2)/(3)m,(1)/(3)m,(1)/(6)m`D. `(1)/(3)m,(2)/(3)m,(2)/(3)m`
Answer» Correct Answer - 4

Discussion

834.	A body of mass m is situated in a potential field `U(x)=U_(0)(1-cosalphax)` when `U_(0)` and `alpha` are constant. Find the time period of small oscialltions.A. `2pisqrt((m)/(U_(0)alpha))`B. `2pisqrt((m)/(U_(0)alpha^(2)))`C. `2pisqrt((m)/(2U_(0)alpha))`D. `2pisqrt((2m)/(U_(0)alpha^(2)))`
Answer» Correct Answer - B Given: `U=U_(0)(1-cosalphax)` `becauseF=-(dU)/(dx)` `thereforeF=-(d)/(dx)[U_(0)(1-cosalphax)]=-U_(0)alphasinalphax` As `alphax` is small, so `sinalphax ~ alphax` `thereforeF=-U_(0)alpha^(2)x` . .. (i) `Fpropx and -ve` shows that F is directed towards the mean position, hence the body executes SHM. For SHM, `F=-kx` Comparing (i) and (ii), we get, `k=U_(0)alpha^(2)` Time period of oscillation, `T=2pisqrt((m)/(k))=2pisqrt((m)/(U_(0)alpha^(2)))`

Discussion

835.	A mass m is vertically suspended from a spring of negligible mass, the system oscillates with a frequency n. what will be the frequency of the system, if a mass `4m` is suspended from the same spring?A. `(n)/(4)`B. `4n`C. `(n)/(2)`D. `2n`
Answer» Correct Answer - C

Discussion

836.	The frequency of oscillation of amass m suspended by a spring is `v_(1)`. If length of spring is cut to one third then the same mass oscillations with frequency `v_(2)`. ThenA. `v_(2) = 3v_(1)`B. `3v_(2)=v_(1)`C. `v_(2)= sqrt(3)v_(1)`D. `sqrt(3)v_(2)=v_(1)`
Answer» Correct Answer - 4

Discussion

837.	If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity isA. `0.10 m//s`B. `0.16 m//s`C. `0.25 m//s`D. `0.5 m//s`
Answer» Correct Answer - B `v_(m)=Aomega` `=50xx10^(-3)xx(2pi)/(2)` `=50xx10^(-3)xx3.14-15.700xx10^(-2)` `=0.157m//s` `~~0.16m//s`

Discussion

838.	When a dampled harmonic oscillator completes 100 oscillations, its amplitude is reduced to `(1)/(3)` of its initial value. When will be its amplitude when it completes 200 oscillations?A. `(1)/(5)`B. `(2)/(3)`C. `(1)/(6)`D. `(1)/(9)`
Answer» Correct Answer - D

Discussion

839.	Mass suspended to a spring is pulled down by 2.5 cm and let go. The mass oscillates with an amplitude ofA. 2.5 cmB. 5 cmC. 7.5 cmD. 10 cm
Answer» Correct Answer - 1

Discussion

840.	The frequency of oscillations of a mass m suspended by spring of `v_(1)`. If the length of the spring is cut to one-third, the same mass oscillates with frequency `v_(2)`. Determing the value of `v_(2)//v_(1)`A. `2`B. `sqrt(2)`C. `4`D. `sqrt(3)`
Answer» Correct Answer - B The frequency of oscillations of mass m and force constant k is, `upsilon_(1)=(1)/(2pi)sqrt((k)/(m))` When the spring is cut to one-half of its length, its force constant is doubled (2k). then frequency of oscillation of mass m will be `upsilon_(2)=(1)/(2pi)sqrt((2k)/(m))" "therefore(upsilon_(2))/(upsilon_(1))=sqrt(2)`

Discussion

841.	A block of mass `M` suspended from a spring oscillates with time period `T`. If spring is cut in to two equal parts and same mass `M` is suspended from one part, new period os oscillation isA. `(T)/(sqrt(2))`B. `sqrt(2)T`C. `T`D. `2T`
Answer» Correct Answer - A `T = 2pi sqrt((m)/(K))`

Discussion

842.	A spring-and-block oscillator with an ideal spring of force constant 180 N/m oscillates with a frequency of 6 Hz. The mass of the block is, approximately,(A) \(\frac18\) kg (B) \(\frac14\) kg (C) 4 kg (D) 8 kg.
Answer» Correct option is (A) \(\frac18\) kg

Discussion

843.	What is meant by periodic and non-periodic motion? Give any two examples, for each motion.
Answer» Periodic motion: Any motion which repeats itself in a fixed time interval is known as periodic motion. Examples: Hands in pendulum clock, swing of a cradle. Non-Periodic motion: Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion. Example: Occurrence of Earth quake, eruption of volcano.

843.

What is meant by periodic and non-periodic motion? Give any two examples, for each motion.

Answer»

Periodic motion: Any motion which repeats itself in a fixed time interval is known as periodic motion.

Examples: Hands in pendulum clock, swing of a cradle.

Non-Periodic motion: Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion.

Example: Occurrence of Earth quake, eruption of volcano.

Discussion

844.	What is meant by Oscillatory motion?
Answer» When an object or a particle moves back and forth repeatedly for some duration of time, its – motion is said to be oscillatory (or vibratory).

Discussion

845.	When a particle is performing linear `SHM` its `KE` is two times its `pE` at a position `A` and its `PE` is two times its `KE` at another position `B`. Find ratio of `KE_(A)` to `KE_(B)`
Answer» `TE_(A) = KE_(A) +PE_(A)` But `KE_(A) = 2PE_(A)`, `TE_(A) = KE_(A) +(1)/(2)KE_(A) = (3)/(2) KE_(A)`. Similarly, `TE_(B) = KE_(B) +PE_(B)`. But `PE_(B) = 2KE_(B)` `TE_(B) = kE_(B) +2KE_(B) = 3KE_(B)`. By the principle of conservation of energy `TE_(A) = TE_(B), (3)/(2) KE_(A) = 3KE_(B)` `(KE_(A))/(KE_(B)) = (2)/(1)`

Discussion

846.	If the KE of a particle performing a SHM of amplitude A is `(3)/(4)` of its total energy, then the value of its displacement isA. `x=pm(A)/(2)`B. `x=pm(A)/(4)`C. `x=pm(sqrt(3A))/(2)`D. `x=pm(A)/(sqrt(2))`
Answer» Correct Answer - A `K.E.=(3)/(4)T.E.` `(1)/(2)momega^(2)(A^(2)-x^(2))=(3)/(4)xx(1)/(2)momega^(2)A^(2)` `A^(2)-x^(2)=(3)/(4)A^(2)` `A^(2)-(3)/(4)A^(2)=x^(2)` `(A^(2))/(4)=x^(2)` `therefore x=(A)/(2)`

Discussion

847.	Variation of acceleration a of a particle executing HSM with displacement x isA. B. C. D.
Answer» Correct Answer - 3

Discussion

848.	A horizontal spring-and-block system consists of a block of mass 1 kg, resting on a frictionless surface, and an ideal spring. A force of 10 N is required to compress the spring by 10 cm. The spring constant of the spring is (A) 100 N.m-1 (B) 10N.m-1 (C) N.m-1 (D) 0.1 N.m-1
Answer» Correct option is (C) N.m^-1

Discussion

849.	At what displacement is the `KE` of a particle performing `SHM` of amplitude `10cm` is three times its `PE`?A. `2.5cm`B. `5cm`C. `7.5 cm`D. `10cm`
Answer» Correct Answer - B `K.E. = 3.P.E.` `P.E. = (1)/(2) m omega^(2)x^(2), K.E. =(1)/(2) m omega^(2) (A^(2)-x^(2))`

Discussion

850.	The `KE` and `PE` , at is a particle executing `SHM` with amplitude `A` will be equal when its displacement isA. `A sqrt(2)`B. `(A)/(sqrt(2))`C. ` (A)/(2)`D. `A`
Answer» Correct Answer - 2

Discussion

Explore topic-wise InterviewSolutions in .

A particle executing SHM has a frequency of 10 Hz when it crosses its equilibrium position with a velocity of `2pi m//s`. Then the amplitude of vibration isA. 0.1 mB. 0.2 mC. 0.4 mD. 1 m

A simple harmonic motion is represented by x(t) = 10 sin(20t + 0.5)Find its amplitude, frequency and initial phase.

A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre isA. `sqrt(2) pi`B. `16 sqrt(2) pi`C. `24 sqrt(2) pi`D. `32 sqrt(2)//pi`

Time period of a particle executing SHM is 16s.At time `t = 2s`, it crosses the mean position . Its amplitude of motion is `( 32sqrt(2))/(pi) m`. Its velocity at `t = 4s` isA. `1 ms^(-1)`B. ` 2ms^(-1)`C. ` 4ms^(-1)`D. `8 ms^(-1)`

The potential energy of a particle oscillating on x-axis is given as `U +20 +(x-2)^(2)`. The mean position is atA. `x =2m`B. `x =1m`C. `x =3m`D. `x =4m`

A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre isA. `sqrt(2)pi`B. `16sqrt(2)pi`C. `32sqrt(2)//pi`D. `4//pi`

A simple pendulum of length `l` is connected to the ceiling of a vehicle that is moving down along a smooth inclined plane `4` in `5`. Then its period of oscillation isA. `2pi sqrt((5l)/(4g))`B. `2pisqrt((4l)/(5g))`C. `2pisqrt((5l)/(3g))`D. `2pi sqrt((3l)/(5g))`

The relation between acceleration and displacement of four particles are given below. The particle undergoing SHM is:A. `a_(x) = +2x`B. `a_(x) =+2x^(2)`C. `a_(x) =- 2x^(2)`D. `a_(x)= -2x`

A block is kept on a rough horizontal plank. The coefficient of friction between block and plank is `1//2`. Plank is undergoing `SHM` is angular frequency `10 rad//s`. Find the maximum amplitude of plank in which the block does slip over plank `(g = 10 m//s^(2))`.

Give some practical examples of S. H. M?

In 20 s, two simple pendulums, P and Q, complete 9 and 7 oscillations, respectively, on the Earth. On the Moon, where the acceleration due to gravity is \(\frac16\)th that on the Earth, their periods are in the ratio (A) 8 : 1 (B) 9 : 7 (C) 7 : 9 (D) 3 : 14.

A seconds pendulum is suspended in an elevator moving with a constant speed in the downward direction. The periodic time (T) of that pendulum is (A) less than two seconds (B) equal to two seconds (C) greater than two seconds (D) very much greater than two seconds.

If T is the time period of a simple pendulum in an elevator at rest, its time period in a freely falling elevator will be(A) \(\frac{T}{\sqrt2}\)(B) √2T (C) 2T (D) infinite.

STATEMENT-1`:` Time periodof oscillation of a simple pendulum mounted in a cabin that is freely falling is zero and STATEMENT -2 `,` In the cabin falling freely under gravity the pendulum is in state of weightlessness.

Time period of a simple pendulum in a freely falling lift will beA. FiniteB. InifiniteC. ZeroD. All of these

The displacement of a particle making S.H.M. is given by `x=6cos(100t+(pi)/(4))m` then the frequency isA. `1.592 Hz`B. `15.92 Hz`C. `159.2 Hz`D. `1592 Hz`

A particle oscillating under a force `vecF=-kvecx-bvecv` is a (k and b are constants)A. Linear oscillationB. Forces oscillationsC. Damped oscilationD. SHM

The displacement of a particle of mass 100 g executing SHm is given by the equation y = 0.2 cos `(100t+(pi)/(4))` meter. Its total energy isA. 1 JB. 10 JC. 2 JD. 20 J

Displacement of a particle executing SHM s `x= 10 ( cos pi t + sin pi t)`. Its maximum speed isA. `5 pi m//s`B. ` 10 pi m//s`C. ` 10 sqrt(2) pi m//s`D. ` 5 sqrt(2 ) pi m//s`

Displacement time equation of a particle executing `SHM` is, `x = 10 sin ((pi)/(3)t+(pi)/(6))cm`. The distance covered by particle in `3s` isA. `5cm`B. `20cm`C. `10cm`D. `15cm`

Displacement of a particle executing SHM s `x= 10 ( cos pi t + sin pi t)`. Its maximum speed isA. 10 cmB. 20 cmC. `5 sqrt(2)` cmD. 50 cm

Time period of a simple pendulum will be double, if weA. decrease the length 2 timesB. decrease the length 4 timesC. increase the length 2 timesD. increase the length 4 times

The velocity of a particle performing simple harmonic motion, when it passes through its mean position iA. infinityB. zeroC. minimumD. maximum

Two springs, of spring constants `k_(1)` and `K_(2)`, have equal highest velocities, when executing SHM. Then, the ratio of their amplitudes (given their masses are in the ratio `1:2`) will beA. `((k_(2))/(k_(1)))^(1//2)`B. `((k_(1))/(k_(2)))^(1//2)`C. `(k_(1))/(k_(2))`D. `k_(1)k_(2)`

Two springs of constants `k_1` and `k_2` have equal maximum velocities, when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will beA. `(k_(1))/(k_(2))`B. `(k_(2))/(k_(1))`C. `((k_(1))/(k_(2)))^(1//2)`D. `((k_(2))/(k_(1)))^(1//2)`

If the earth were a homogeneous sphere of radius R and a straight hole bored in it through its centre, show that a body dropped into the hole will execute SHM and find its time period.

The total energy of a simple pendulum is x. When the displacement is half of amplitude,its KE will beA. `(x)/(2)`B. `(x)/(4)`C. `(3x)/(4)`D. `x`

Two identical springs have the same force constant of `147Nm^(-1)`. What elongation will be produced in each spring in each case shown in figure? `g=9.8ms^(-2)`. A. `(1)/(6)m,(2)/(3) m, (1)/(3)m`B. `(1)/(3)m,(1)/(3)m,(1)/(3)m`C. `(2)/(3)m,(1)/(3)m,(1)/(6)m`D. `(1)/(3)m,(2)/(3)m,(2)/(3)m`

A body of mass m is situated in a potential field `U(x)=U_(0)(1-cosalphax)` when `U_(0)` and `alpha` are constant. Find the time period of small oscialltions.A. `2pisqrt((m)/(U_(0)alpha))`B. `2pisqrt((m)/(U_(0)alpha^(2)))`C. `2pisqrt((m)/(2U_(0)alpha))`D. `2pisqrt((2m)/(U_(0)alpha^(2)))`

A mass m is vertically suspended from a spring of negligible mass, the system oscillates with a frequency n. what will be the frequency of the system, if a mass `4m` is suspended from the same spring?A. `(n)/(4)`B. `4n`C. `(n)/(2)`D. `2n`

The frequency of oscillation of amass m suspended by a spring is `v_(1)`. If length of spring is cut to one third then the same mass oscillations with frequency `v_(2)`. ThenA. `v_(2) = 3v_(1)`B. `3v_(2)=v_(1)`C. `v_(2)= sqrt(3)v_(1)`D. `sqrt(3)v_(2)=v_(1)`

If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity isA. `0.10 m//s`B. `0.16 m//s`C. `0.25 m//s`D. `0.5 m//s`

When a dampled harmonic oscillator completes 100 oscillations, its amplitude is reduced to `(1)/(3)` of its initial value. When will be its amplitude when it completes 200 oscillations?A. `(1)/(5)`B. `(2)/(3)`C. `(1)/(6)`D. `(1)/(9)`

Mass suspended to a spring is pulled down by 2.5 cm and let go. The mass oscillates with an amplitude ofA. 2.5 cmB. 5 cmC. 7.5 cmD. 10 cm

The frequency of oscillations of a mass m suspended by spring of `v_(1)`. If the length of the spring is cut to one-third, the same mass oscillates with frequency `v_(2)`. Determing the value of `v_(2)//v_(1)`A. `2`B. `sqrt(2)`C. `4`D. `sqrt(3)`

A block of mass `M` suspended from a spring oscillates with time period `T`. If spring is cut in to two equal parts and same mass `M` is suspended from one part, new period os oscillation isA. `(T)/(sqrt(2))`B. `sqrt(2)T`C. `T`D. `2T`

A spring-and-block oscillator with an ideal spring of force constant 180 N/m oscillates with a frequency of 6 Hz. The mass of the block is, approximately,(A) \(\frac18\) kg (B) \(\frac14\) kg (C) 4 kg (D) 8 kg.

What is meant by periodic and non-periodic motion? Give any two examples, for each motion.

What is meant by Oscillatory motion?

When a particle is performing linear `SHM` its `KE` is two times its `pE` at a position `A` and its `PE` is two times its `KE` at another position `B`. Find ratio of `KE_(A)` to `KE_(B)`

If the KE of a particle performing a SHM of amplitude A is `(3)/(4)` of its total energy, then the value of its displacement isA. `x=pm(A)/(2)`B. `x=pm(A)/(4)`C. `x=pm(sqrt(3A))/(2)`D. `x=pm(A)/(sqrt(2))`

Variation of acceleration a of a particle executing HSM with displacement x isA. B. C. D.

A horizontal spring-and-block system consists of a block of mass 1 kg, resting on a frictionless surface, and an ideal spring. A force of 10 N is required to compress the spring by 10 cm. The spring constant of the spring is (A) 100 N.m-1 (B) 10N.m-1 (C) N.m-1 (D) 0.1 N.m-1

At what displacement is the `KE` of a particle performing `SHM` of amplitude `10cm` is three times its `PE`?A. `2.5cm`B. `5cm`C. `7.5 cm`D. `10cm`

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