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751.

A body `A` of mass moving with velocity `v` while passing through its mean position collides in perfect inelastically with a body `B` of same mass which is connected to a vertical wall through a spring whose spring constant is `k`. After collision it sticks to `B` and executes `S.H.M.` Find the amplitude of reulting motion:A. `sqrt((m)/(v))v`B. `sqrt((m)/(2k))v`C. `(m)/(k)sqrt(v)`D. `(m)/(2k)sqrt(v)`

Answer» Correct Answer - B
Assuming the collision lests for a small interval. We can apply conservation of momentum and get the common velocity = `(v)/(2) K.E. =(1)/(2) (2m) ((v)/(2))^(2) = (1)/(4)mv^(2)`
This is also the total energy of system as the spring is unstretched at this moment. If the amplitude is `A`, total energy `=(1)/(2) kA^(2) :. (1)/(2) kA^(2) = (1)/(4)mv^(2)`
`:.A = sqrt((m)/(2k)).v`
752.

In an experiment with bar pendulum having four holes, the same time period is recorded when it is suspended at distances `21cm, 24cm, 40cm` nad `52cm` respectively form one end. The length of bar pendulum isA. `84 cm`B. `72 cm`C. `64 cm`D. `60 cm`

Answer» Correct Answer - C
Holes should be at equal distance from either end to have same time period.
753.

A sphere of radius r is kept on a concave mirror of radius of curation `R`. The arrangement is kept on a horizontal surface (the surface of concave mirror is friction less and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes `S.h.M.` The period of oscillation will beA. `2pi sqrt([(R-r)1.4//g])`B. `2pi sqrt([(R-r)//g])`C. `2pi sqrt([(Rr//g)])`D. `2pi sqrt([(R//gr)])`

Answer» Correct Answer - A
`T = 2pi (sqrt(((R-r)(1+beta))/(g)))`
754.

A rod with rectangular cross section oscillates about a horizontal axis passing through one of its ends and it behaves like a seconds pendulum, its length will beA. `1.5 m`B. `1m`C. `3m`D. `2m`

Answer» Correct Answer - A
`T = 2pi sqrt((k^(2)/(l)+1)/(g))`here `l = (L)/(2) & K^(2) = (L^(2))/(12)`
755.

Under what conditions can we consider the oscillations of a simple pendulum to be linear simple harmonic?

Answer»

The oscillations of a simple pendulum are approximately linear simple harmonic only if 

1. the amplitude of oscillation is very small compared to its length 

2. the oscillations are in a single vertical plane.

756.

A simple pendulum of length 121 cm takes 11/3 minutes to execute 100 oscillations. Then the time that another simple pendulum of length 81 cm takes to execute the same number of oscillations isA. 2 minutesB. 3 minutesC. 6 minutesD. 9 minutes

Answer» Correct Answer - B
`t_(1)=nT_(1)andt_(2)=nT_(2)`
`(t_(2))/(t_(1))=(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt((81)/(21))`
`(t_(2))/(t_(1))=(9)/(11)`
`therefore t_(2)=(9)/(11)xx(11)/(3)=3` minutes.
757.

Starting from the extreme position, the time taken by an ideal simple pendulum to travel a distance of half the amplitude isA. `T//6`B. `T//12`C. `T//13`D. `T//4`

Answer» Correct Answer - A
`t=(T)/(12)`
758.

A point mass oscillates along the x-axis according to the law `x=x_(0) cos(moegat-pi//4). If the acceleration of the particle is written as `a=A cos(omegat+delta), the .

Answer» Velocity `v = (dx)/(dt) =- x_(0) omega sin (omegat -(pi)/(4))`
Acceleration `a = (dv)/(dt) =- x_(0) omega^(2) cos (omegat -(pi)/(4))`
`a = x_(0) omega^(2) cos(omegat+(3pi)/(4)) ….(1)`
Given that `a = A cos (omegat +delta) ….(2)`
from (1) & (2), `A = x_(0) omega^(2), delta = (3pi)/(4)`
759.

A simple pendulum oscillates with an amplitude of `2xx10^(-2)m`. If the force acting on it at extreme position is 4 N. Then the time that another simple pendulum of length 81 cm takes to execute the same number of oscillations isA. 4NB. 3NC. 2ND. 1N

Answer» Correct Answer - C
`(F_(m))/(F_(e))=(kx_(m))/(kx_(e))`
`=((A)/(2))/(A)=(1)/(2)`
`F_(m)=(F_(e))/(2)=(4)/(2)=2N`
760.

Two identical blocks `P` and `Q` have mass `m` each. They are attached to two identical springs initially unstretched. Now the left spring (along with `P)` is compressed by `A//2` and the right spring (along with `Q`) is compressed by `A`. Both the blocks are released simultaneously. They collide perfectly inelastically. Initially time period of both the blocks was `T`. The time period of oscillation of combined mass isA. `(T)/(sqrt(2))`B. `sqrt(2)T`C. `T`D. `(T)/(2)`

Answer» Correct Answer - C
They will collide at their positions because time period of both are same and that is `2pi sqrt((m)/(k))`. After collision combined mass is `2m` and `K_(eff) =2K`. Hence, time period remains unchanged.
761.

A particle oscillates about the stable equilibrium position `x_(0)` with a small amplitude subjected to a force that has an associated potential energy `u(x)`. Which of the following about `u(x)` is true (no implsive force)A. `u(x)` must be symmetric about `x_(0)`B. `u(x)` must have a minima at `x_(0)`C. `u(x)` may have maxima at `x_(0)`D. `u(x)` may be position in the vicinity of `x_(0)`

Answer» Correct Answer - B::D
The only need for oscillation is potential energy should have a minima and for the amplitude given it should not cross the valley.
762.

A cube made of wood having specific gravity `0.4` and side length `a` is floated in a large tanck full of water. Which action woukd change the depth to which block is submerged? `(gamma_("wood") lt gamma_("water"))`A. more water is added in the tankB. atmospheric pressure increasesC. the tank is accelerated upwardsD. A small coin is placed over the cube

Answer» Correct Answer - D
`mg = F_(b), a^(2) xx 0.4 rho g = a^(2) h rho g`
`h = (2)/(5)a, m = a ^(3) xx 0.4, F_(net) =a^(2) x rho g`
`T = 2pi sqrt((a^(3) xx 0.4 rho)/(a^(2) rho g)) , T = 2pi sqrt((2a)/(5g))`
763.

Two simple pendulums of length `100m` and `121m` start swinging together. They will swing together again afterA. the longer pendulum makes `10` oscillationsB. the shorter pendulum makes `10` oscillationsC. the longer pendulum makes `11` oscillationsD. the shorter pendulum makes `20` oscillations

Answer» Correct Answer - A
`nT_(s) = (n-1)T_(1)`
764.

A body having netural frequency `v_(o)` is executing forced oscillations under driving force of frequency `v`. The system will vibrateA. With the frequency of driving force `v`B. With netural frequency `v_(o)`C. With mean frequency `((v_(o)+v)/(2))`D. None of the above

Answer» Correct Answer - A
765.

Two simple pendulum first of bob mass `M_(1)` and length `L_(1)` second of bob mass `M_(2)` and length `L_(2) M_(1) = M_(2)` and `L_(1)` = 2L_(2)`. if the vibrational energy of both is same which is correct?A. Amplitude of B greater than AB. Amplitude of B smaller than AC. Amplitude will be sameD. None of the above

Answer» Correct Answer - B
Frequency, `n = (1)/(2pi)sqrt((g)/(l)) or n prop (1)/(sqrt(l))`
`therefore" "(n_(1))/(n_(2))= sqrt((l_(2))/(l_(1)))=sqrt((L_(2))/(2L_(2)))=(1)/(sqrt(2))=n_(2)=sqrt(2) n_(1)`
`rArr" "n_(2) gt n_(1)`
Energy, `E = (1)/(2)m omega^(2)a^(2)=2pi^(2)mn^(2)a^(2)`
and `a^(2)prop (1)/(mn^(2))rArr (a_(1)^(2))/(a_(2)^(2))=(m_(2)n_(2)^(2))/(m_(1)n_(1)^(2))" "(because "E is same")`
Given, `n_(2) gt n_(1) and m_(1) = m_(2)`
`rArr" "a_(1) gt a_(2)`. So, amplitude of B smaller than A.
766.

To start a pendulum swinging, usually you pull it slightly to one side and release. What kind of energy is transferred to the mass in doing this?

Answer»

On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.

767.

There is a spring with netural length `L_(0)`. Two masses `m_(1)` and `m_(2)` are connected to both of its ends as shown in figure. The whole system is held at rest. At any time `t =0, m_(2)` is released and system starts free fall. Initial stretched length of spring before fall is `L.` what is the displacement of centre of mass as function of time? A. `g t^(2)`B. `(1)/(2)g t^(2)`C. `(g)/(K)t^(2)`D. `(m_(1)+m_(2))/(m_(1)m_(2))xxt`

Answer» Correct Answer - B
`a_(cm) = (m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2)), a_(1) =a_(2) = g`
`:. S = (1)/(2) a_(cm)t^(2) = (1)/(2)g t^(2)`
768.

A pendulum of mass `m` and length `L` is connected to a spring as shown in figure. If the bob is displaced slightly from its mean position and released, it performs simple harmonic motion. The angular frequency of the oscillation of bob is A. `sqrt((g)/(L))`B. `sqrt((Kh^(2)+mgL)/(mL^(2)))`C. `sqrt(3g)/(L)`D. `sqrt((Kh^(2))/(mL^(2)))`

Answer» Correct Answer - B
`tau_(rest) =- T alpha` For small angular displacement `theta mgl sin theta = kyh =- I alpha sin theta = theta, tan theta = (y)/(h) = theta`
769.

The maximum potential energy (PE) of a particle in SHM is 2 × 10-4 J. What will be the PE of the particle when its displacement from the mean position is half the amplitude of SHM ?

Answer»

(PE)max = 1/2 kA2 , PE = 1/2 kx2 

∴ PE = (PE)max (x/A)2 = 2 × 10-4 J × (x/A)2

= 5 × 10-5 J is the required answer.

770.

A particle under the action of a force has a period of `3s` and under the action of another force it has a period `4sec` in `SHM`. What will be its period under the combined action of both forces in the same direction?A. `7s`B. `5s`C. `2.4s`D. `0.4s`

Answer» Correct Answer - C
`T = (T_(1)T_(2))/(sqrt(T_(1)^(2)+T_(2)^(2)))`
771.

A particle executes `SHM` with an amplitude of `10cm` and frequency `2 Hz`. At `t = 0`, the particle is at a point where potential energy and kinetic energy are same. The equation for its displacement isA. `x = 0.1 sin (4pi t+(pi)/(4))m`B. `x = 0.1 (sin 4pi t)m`C. `x = 0.1 cos(4pi t+(pi)/(3))`D. `x = 0.1(sin4pit-(pi)/(3))m`

Answer» Correct Answer - A
`x =A sin (omegat +phi), phi = (pi)/(4)`
772.

What do you mean by amplitude of the simple harmonic motion?

Answer»

Maximum displacement of the particle is called the amplitude of the S.H. motion.

773.

How would the time period of spring mass system change, when it is made to oscillate horizontally and then vertically?

Answer»

The time period remains the same in both the cases.

774.

A body of mass `1kg` is suspended from a weightless spring having force constant `600N//m`. Another body of mass `0.5 kg` moving vertically upward hits the suspended body with a velocity of `3.0m//s` and get embedded in it. Find the frequency of oscillations and amplitude of motion.A. `(pi)/(10)Hz`B. `(10)/(pi)Hz`C. `(1)/(2pi)Hz`D. 3.142 Hz

Answer» Correct Answer - B
`n=(1)/(2pi)sqrt((k)/(m))`
`=(1)/(2pi)sqrt((600)/(1.5))=(1)/(2pi)xx20=(10)/(pi)Hz`
775.

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
Given, `y_(1)=a sin omega t and y_(2)=b cos omega t = b sin (omega t + (pi)/(2))`
The resultant displacement is given by
`y = y_(1)+y_(2)=sqrt(a^(2)+b^(2))sin (omega t + phi)`
Hence, the motion of superimposed wave is simple harmonic with amplitude `sqrt(a^(2)+b^(2))`
776.

A simple harmonic oscillation is represented by the equation. `y=0.40sin(440t+0.61)` Here y and t are in m are s respectively. What are the values of (i) amplitude (ii) angular frequency (iii) frequency of oscillations (iv) time period of oscillations and (v) initial phase?

Answer» Correct Answer - (i). 0.40m
(ii). 440 rad `s^(-1)`
(iii) 70 Hz
(iv). 0.0143s
(v). 0.61 rad
777.

Write the displacement equation representing the following conditions obtained for a simple harmonic motion: amplitude=0.01m, frequency=600Hz initial phase `=pi//6`

Answer» Correct Answer - `y=0.01sin(1200pit+pi//6)`
778.

A body oscillates with SHM, accroding to the equation, `x=(5.0m)cos[(2pirads^(-1))t+pi//4]` At `t=1.5s`, calculate the `(a)` diplacement `(b)` speed and `(c)` acceleration of the body.

Answer» The angular frequency `omega` of the body `= 2pis^(-1)` and its time period `T = 1s`
At `t=1.5s`
(a) displacement `= (5.0 m) cos [(2 pi s^(-1))xx1.5 s +pi//4]`
`=(5.0 m) cos [(3pi + pi//4)]`
`= -5.0 xx 0.707 m`
`=-2.535` m
(b) Using Eq. (14.9), the speed of the body
`= -(5.0 m)(2pi s^(-1))sin [(2pi s^(-1))xx1.5 s + pi//4]`
`=-(5.0 m)(2pi s^(-1))sin [(3pi +pi//4)]`
`=10 pi xx 0.707 ms^(-1)`
`=22 ms^(-1)`
(c) Using Eq. (14.10), the acceleration of the body
`= (2pi s^(-1))^(2) xx` displacement
`= -(2pi s^(-1))^(2) xx (-3.535m)`
`= 140 ms^(-2)`.
779.

A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the particle at the time `T//12` is: `(T="time period")`A. `1 : 4`B. `2 : 1`C. `3 : 1`D. `4 : 1`

Answer» Correct Answer - C
When `t = (T)/(12)`, then `x = A "sin"(2pi)/(T)xx(T)/(12)=(A)/(2)`
`KE = (1)/(2)mv^(2)=(1)/(2)m omega^(2)(r^(2)-x^(2))`
`= (1)/(2)m omega^(2)(A^(2)-(A^(2))/(4))`
`=(3)/(4)((1)/(2)momega^(2)A^(2))`
`PE = (1)/(2)m omega^(2)x^(2)=(1)/(4)((1)/(2)m omega^(2)A^(2))`
`(KE)/(PE)=(3)/(1)`
780.

The periodic time of a body executing S.H.M. is 2 sec. after how much interval from t = 0., will its displacement be half of its amplitude?

Answer»

Here, T = 2s; t = ?, y = a/2

Now, y = asin ωt = asin\(\frac{2\pi}{T}t\) (∵ T = 2s)

∴  \(\frac{a}{2}=asin\frac{2\pi}{T}t=asin\,\pi t\)

Or  sin πt = \(\frac{1}{2}\)

∴ sin 30º = sin \(\frac{\pi}{6}\) 

Or πt = \(\frac{\pi}{6}\)  or t = \(\frac{1}{6}\)sec. 

781.

Velocity and displacement of a body executing S.H.M. are out of phase by π/2. Why?

Answer»

Let displacement on y-axis is given by

y = asin ωt

Where a is the amplitude, ω is the angular velocity and  t is the instantaneous time.

Then velocity is given by

v = \(\frac{dy}{dt}\) = aω cos aω = aω sin ( \(\frac{\pi}{2}\) + aω)

The above equation clearly show that displacement and velocity of a body executing S.H.M. are out of phase by \(\frac{\pi}{2}\).

782.

The displacement of an elastic wave is given the function y = 3 sin ωt + 4 cos ωt.When y is in cm and t is in second. Calculate the resultant amplitude.

Answer»

y =3sin ωt +4 cos ωt …(i)

Let, 3 = a cos ϕ …(ii)

4 = a sin ϕ …(iii)

∴ = a cos ϕ sin ωt + a sinϕ cos ωt …(iv)

y = a sin(ωt + φ)

From eq. (s) (i) & (iii)-

tan ϕ = \(\frac{3}{4}\) or ϕ = tan-1\(\frac{3}{4}\) 

Squaring (ii) & (iii), then adding- 

a2cos2ϕ+ a2sin2ϕ = 32 + 42

a2(cos2ϕ + sin2ϕ) = 9 + 16

a2 = 25

or a = 5 or amplitude, = 5 cm

∴ y1 = 5 sin (ωt + ϕ), ∵ ϕ = tan-1\(\frac{4}{3}\) 

Hence, new amplitude, a = 5cm.

783.

STATEMENT-1`:` Time period of a simple pendulum changeswhen the solid bob is replaced by a hollow sphere of same radius but difference mass and STATEMENT-2 `:` The time period of a simple pendulum depends on force acting on bob per unit mass dueto the earth.

Answer» Correct Answer - 4
784.

The period of a simple pendulum is found to be increased by `50%` when the length of the pendulum is increased by `0.6m`. The initial length isA. 0.48 mB. 0.5 mC. 0.994 mD. 0.3 m

Answer» Correct Answer - A
`T_(2)=(1.5)T_(1)andl_(2)=l_(1)+0.6`
`(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt((l_(1)+0.6)/(l_(1)))therefore (1.5)^(2)=(l_(1)+0.6)/(l_(1))`
`therefore 2.25=(l_(1)+0.6)/(l_(1)) therefore 1.25l_(1)=0.6`
`l_(1)=(0.6)/(1.25)=(0.6xx4)/(5)=(2.4)/(5)`
`=0.48m`
785.

Simple pendulum is executing simple harmonic motion with time period `T`. If the length of the pendulum is increased by `21%`, then the increase in the time period of the pendulum of the increased length is:A. 0.1 sB. 0.2 sC. 0.4 sD. 0.15 s

Answer» Correct Answer - B
`(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt(1.21)`
`therefore (T_(2))/(T_(1))=1.1`
`(T_(2)-T_(1))/(T_(1))=0.1`
`therefore T_(2)-T_(1)=0.1xx2=0.2s`
786.

A pendulum suspended from the ceiling of the train has a time period of two seconds when the train is at rest, then the time period of the pendulum, if the train accelerates 10 `m//s^(2)` will be `(g=10m//s^(2))`A. 2sB. `2 sqrt(2) s`C. `2//sqrt(2) s`D. `2^(3//4)`

Answer» Correct Answer - D
The net acceleration = `sqrt((a^(2)+g^(2)))`
`=sqrt(100+100)=sqrt(200)`
`T_(1)=2pisqrt((l)/(g))and T_(2)=2pisqrt((l)/(a_(R)))`
`(T_(2))/(T_(1))=sqrt((10)/(10sqrt(2)))therefore T_(2)=(2)/(sqrt(2))`
787.

The time period of oscillation of a simple pendulum is `sqrt(2)s`. If its length is decreased to half of initial length, then its new period isA. `1s`B. `0.707s`C. `0.414s`D. `0.5s`

Answer» Correct Answer - A
`Talpha sqrt(l)`
788.

A train moving at a uniform speed has a simple pendulum hung from the ceiling of one of the compartments. The train suddenly experiences a uniform acceleration, during this interval, the time period of the pendulumA. remains sameB. decreasesC. increasesD. becomes infinite

Answer» Correct Answer - B
789.

Tension in the string is minimum when the pendulum is atA. mean positionB. extreme positionC. mid way between mean and extreme positionD. any position

Answer» Correct Answer - B
790.

IF the length of a simple pendulum is increased, its maximum velocity during oscillationA. decreasesB. increasesC. remains sameD. is zero

Answer» Correct Answer - A
791.

The tension in the string of a simple pendulum isA. constantB. maximum in the extreme positionC. zero in the mean positionD. minimum at extreme position

Answer» Correct Answer - D
792.

In SHM, graph of which of the following is a straight line?A. T.E. against displacementB. P.E. against displacementC. acceleration against timeD. velocity against displacement

Answer» Correct Answer - A
793.

A pendulum is making one oscilla¬tion in every two seconds. What is the frequency of oscillation?

Answer»

f = \(\frac{1}{T}\) = \(\frac{1}{2}\) s-1.

794.

When is the tension maximum in the spring of a simple pendulum?

Answer»

Mean position.

795.

State force law for a SHM.

Answer»

Force, F = mω2

m = mass 

ω = angular speed 

x = displacement

796.

What Is the phase difference between the displacement and velocity in a SHM?

Answer»

90° (π/2 radians)

797.

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period would beA. `(pi)/(100)s`B. `(pi)/(50)s`C. `(pi)/(20)s`D. `(pi)/(10)s`

Answer» Correct Answer - A
`TE =PE +KE`
798.

A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3s. The time period is(a) 15s (b) 6s (c) 12s (d) 9s

Answer»

(c) 12s 

Time period of Oscillation = 2 × (time taken to go from A to B + the next time taken to return at B) 

= 2 × (3 + 3) 

= 2 × 6 

Time period = 12s

799.

The displacement `x` of a particle in motion is given in terms of time by `x(x-4) =1 -5 cos omegat`A. the particle executes `SHM`B. the particle executes oscillatory motion which is not `SHM`C. the motion of the particle is neither oscillatory nor simple harmonicD. the particle is not acted upon by a fprce when it is at `x = 4`

Answer» Correct Answer - A
`(d^(2)(x-2))/(dt^(2)) = (-omega^(2))/(4)(x-2)`
`:.` The particle executes `SHM`
800.

The period of a particle in SHM is `8s`. At `t=0` it is at the mean position. The ratio of the distances travled by it in the first and the second second isA. `(1)/(2)`B. `(1)/(sqrt(2))`C. `sqrt(2)`D. `sqrt(2)+1`

Answer» Correct Answer - D
`y_(1) =Asin ((2pi)/(T)xxt_(1)), y_(2) = A sin ((2pi)/(T)xx(t_(1)+t_(2)))` Distance travelled in second, `y_(2)^(1) = y_(2) - y_(1)`