1003 + Interview Questions in SIMPLE-HARMONIC-MOTION IN GENERAL AWARENESS Page 10 InterviewSolution

451.	An block rides on position that is moving vertically with simple harmonic motion. The maximum speed of the piston is 2m/s. At what minimum amplitude of motion will the block and piston separate? (g=10m/`s^(2)`)A. 20 cmB. 30cmC. 40 cmD. 50 cm
Answer» Correct Answer - C

Discussion

452.	The vertical motion of a ship at sea is described by the equation `(d^(x))/(dt^(2))=-4x` , where x is the vertical height of the ship (in meter) above its mean position. If it oscillates through a height of 1 mA. its maximum vertical speed will be `1 ms^(-1)`B. its maximum vertical speed will be `2 ms^(-1)`C. its greater vertical acceleration is `2 ms^(-2)`D. its greater vertical acceleration is `1 ms^(-2)`
Answer» Correct Answer - B (b) Comparing with `(d^(2)x)/(dt^(2))=-omega^(2)x` we have , `omega=2 "rad"s^(-1)` `v_("max")=omegaA=2ms^(-1)` `a_("max")=omega^(2)A=4 ms^(-2)`

Discussion

453.	A body undergoing `SHM` about the origin has its equation is given by `X=0.2 cos 5pit`. Find its average speed from `t=0 to t=0.7 sec`.
Answer» Correct Answer - `2 m//s`

Discussion

454.	The acceleration-displacement `(a-X)` graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscillation.
Answer» Correct Answer - `(1)/(2pi)sqrt((beta)/(alpha))`

Discussion

455.	Two non - viscous, incompressible and immiscible liquids of densities (rho) and (1.5 rho) are poured into the two limbs of a circular tube of radius ( R) and small cross section kept fixed in a vertical plane as shown in fig. Each liquid occupies one fourth the cirumference of the tube. . (a) Find the angle (theta) that the radius to the interface makes with the verticles in equilibrium position. (b) If the whole is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the period of these oscillations.
Answer» Correct Answer - A::B::C At equilibrium taking of liquid about `O` (Torque due to liquid densith `rho` ) = )Torque due to loquid of density `1.5 rho`) `m_(2)g xx QM =m_(1)g xx PN` `:. M_(2)g sin (45^(@)) =m_(1)g R sin(45^(@)-theta)` `V rho g R sin (45^(@)+theta) = 1.5 V rho g R sin(45^(@)-theta)` ...(i) ltbr `implies (SIN(45+theta))/(sin(45-theta)) =1.5` `implies (sin45^(@)cos theta + cos 45^(@) sin theta)/(sin45^(@)cos theta - cos 45^(@) sin theta)=3/2` `implies tan theta = 1/5` `tau = m_(2)g R sin(45^(@)+theta+alpha)-m_(1)g R sin gR sin(45^(@)-theta-alpha)` `=V rho g R sin(45^(@)-theta - alpha)-1.5 V rho g R(45^(@)-theta-alpha)` `=V rho g Rsin(theta+45^(@)) cos alpha+V rho g R cos(45^(@)+theta)` `sin alpha-1.5 V rho g R sin(45^(@)-theta) cos alpha + 1.5 V rho R cos (45^(@)-theta) sin alpha` Using eq. (i) we get `tau = V rho g R[cos (45^(@)+theta)sin alpha+1.5 cos (45^(@)-theta)sin alpha]` `tau = V rho g R[cos (45^(@)+theta)+1.5 cos (45^(@)-theta)sin alpha] When `alpha` is small (given) `:. sin alpha ~~ alpha` `tau = V rho g R[cos(45^(@)+theta)+1.5 cos (45^(@)-theta)]alpha` Since, `tau` and `alpha` are proportional and directed towards mean position. `:. The motion is simple harmonic. Moment of interia about `O` is `I = V rhoR^(2) +1.5V rho R^(2)` `T = 2 pi sqrt((1)/(C)` `=2 pi (sqrt((V rho xx 2.5 R^(2))))/(sqrt([cos(45+theta)+1.5 cos (45-theta)]Vrho gR))` `=2pi (sqrt(1.803R))/(sqrt(g))` (using the value ` tan theta = 1/5`).

Discussion

456.	Which of the following statements is wrong?A. De-Brogile wavelenfth are probability waves and there is no physical existenece of these.B. De-Brogile wavelength of a moving particle is inversely proportional to its momentum.C. Wave nature is associated with atomic particles only.D. In general wave nature of matter is not observed.
Answer» Correct Answer - C De brogile waves are probability waves and are applicable for all the objects. Wave nature is observed for the small particles like electrons.

Discussion

457.	The de Broglie waves are associated with moving particles. These particle may beA. electronsB. `He^(+), Li^(2+)` ionsC. Cricket ballD. All of the above
Answer» Correct Answer - D De brogile waves are indendent of shape & size of the object.

Discussion

458.	Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)` is close to (Avogadro number `= 6.023 xx 10^(23) // "g. mole"`) :-A. `7.5 xx 10^(21)`B. `1.75 xx 10^(22)`C. `6.2 xx 10^(21)`D. `1.25 xx 10^(22)`
Answer» Correct Answer - A No. of nuclei remains at any instant `N= N_(0)/(2^(t//T_(1//2)))=N_(0)/sqrt(2)` `N_("decayed")=N_(0)-N=N_(0) [1-1/sqrt(2)]` `=1/24xx6.023xx10^(23)xx0.3` `~~7.5xx10^(21)`

Discussion

459.	Two particles are executing S.H.M. The equation of their motion are `y_(1) = 10 sin (omega t + (piT)/4),y_(2)=25 sin (omegat+(sqrt3piT/4))` . What is the ratio of their amplitudeA. `1:1`B. `2:5`C. `1:2`D. None of these
Answer» Correct Answer - B

Discussion

460.	Two identical small elastic balls have been suspended using two strings of different length (see fig (a)). Pendulum A is pulled to left by a small angle `theta_(0)` and released. It hits ball B head on which swings to angle `2 theta _(0)` from the vertical. Calculate the time period of oscillation of A if its length is known to be L.
Answer» Correct Answer - N//A

Discussion

461.	The angular velocity and amplitude of simple pendulum are `omega` and r respectively. At a displacement x from the mean position, if its kinetic energy is T and potential energy is U, find the ration of T to U.A. `X^2omega^2//(a^2-X^2omega^2)`B. `X^2//(a^2-X^2)`C. `(a^2-X^2omega^2)//X^2omega^2`D. `(a^2-X^2)//X^2`
Answer» Correct Answer - B `T/V=(1/2momega^2(a^2-X^2))/(1/2momega^2X^2)=(a^2-X^2)/(X^2)`

Discussion

462.	The given figure shows the variation of the kinetic energy of a simple pendulum with its angular displacement `(theta)` from the vertical. Mass of the pendulum bob is m = 0.2 kg. Find the time period of the pendulum. Take `g = 10 ms^(-2)`.
Answer» Correct Answer - T=2.80s

Discussion

463.	If the displacement of simple pendulum at any time is `0.02` m and acceleration is `2m//s^(2)`, then in this time angular velocity will beA. 100 rad/sB. 10 rad/sC. 1 rad/sD. `0.1` rad/s
Answer» Correct Answer - B Acceleration `\|alpha\|=omega^(2)x` or `omega=sqrt(alpha)/(x)` `=sqrt((2)/(0.002))=10rad//s`

Discussion

464.	A simple wave motion represented by `y=5(sin 4pi t+sqrt3 cos 4pit)`. Its amplitude isA. 5B. `5sqrt3`C. `10sqrt3`D. 10
Answer» Correct Answer - D `y=5(sin 4pit+sqrt3 cos 4pit)` `y=5(sin 4pi t+5sqrt3 cos 4pi t)` `A=sqrt(A_(1)^(2)+A_(2)^(2))` `A=sqrt((5)^(2)+(5sqrt3)^(2))` `=sqrt(25+75)=sqrt(100)=10`

Discussion

465.	The frequency of a simple pendulum is `n` oscillations per minute while that of another is `(n + 1)` oscillations per minute. The ratio of length of first second is-A. `(n)/(n+1)`B. `(n+(1)/(n))^(2)`C. `((n+1)/(n))^(2)`D. `((n)/(n+1))^(2)`
Answer» Correct Answer - C `T = 1/f = 2pisqrt((l)/(g)) rArr f = (1)/(2pi) sqrt((g)/(l))` `(f_(1))/(f_(2)) = sqrt((l_(2))/(l_(1))) rArr ((n)/(n+1))^(2) = (l_(2))/(l_(1)) rArr = (l_(1))/(l_(2)) = ((n+1)/(n))^(2)`

Discussion

466.	One mass m is suspended from a spring. Time period of oscilation is T. now if spring is divided into n pieces & these are joined in parallel order then time period of oscillation if same mass is suspended.A. `2pisqrt((n^(2)m)/k)`B. `2pisqrt((n^(2)k)/m)`C. `2pisqrt(m/(n^(2)k))`D. `2pisqrt(k/(n^(2)m))`
Answer» Correct Answer - C `T=2pisqrt(m/k)` And spring is devided into n piece and connected in parallel combination then `k_(eq)=n^(2)xxk` So new time period `=2pisqrt(m/(n^(2)k))`

Discussion

467.	In a laboratory experiment with simple pendulum it was found it took 36 s to complete 20 oscillations when the effective length was kept that 80 cm. Calculate the acceleration due to gravity from these data.
Answer» The time period of a simple pendulum is given by `T=2pisqrt(lg^-1)` or `g=(4pi^2l)/T^2` ………….i In tgeh experiment described in the question the time period is `T=(36s)/20=1.8s` Thus by i `g=(4pi^2xx0.80m)/((1.8s)^2)=98.75ms^-2`

Discussion

468.	A simple pendulum has a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of the pendulum willA. Remains unchangedB. IncreaseC. DecreaseD. Become erratic
Answer» Correct Answer - B

Discussion

469.	Two spheres A and B of the same mass m and the same radius are placed on a rough horizontal surface. A is a uniform hollow sphere and B is uniform solid sphere. They are tied centrally to a light spring of spring constant k as shown in figure. A and B are released when the extension in the spring is `x_(0)`. Friction is sufficient and the spheres do not slip on the surface. Find the frequency and amplitude of SHM of the sphere A.
Answer» Correct Answer - `f=(1)/(2pi) sqrt((46k)/(35 g)) ; A_(1)=(21)/(46) x_(0)`

Discussion

470.	A large horizontal turntable is rotating with constant angular speed `omega` in counterclockwise sense. A person standing at the centre, begins to walk eastward with a constant speed V relative to the table. Taking origin at the centre and X direction to be eastward calculate the maximum X co-ordinate of the person.
Answer» Correct Answer - `(V)/(omega)`

Discussion

471.	A spherical cavity of radius`(R )/(2)` is removed from a solid sphere of radius R as shown in fig. The sphere is placed on a rough horizontal surface as shown. The sphere is given a gentle push. Friction is large enough to prevent slippage. Prove that the sphere perform SHM and find the time period.
Answer» Correct Answer - `T=2 pi sqrt((177R)/(10g))`

Discussion

472.	The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are 2 m/s and` 4 m//s^(2)`. Then angular velocity will beA. 3 rad/secB. 0.5 rad/secC. 1 rad/secD. rad/se
Answer» Correct Answer - D

Discussion

473.	Time period of a spring mass system can be changed byA. changing the massB. acclerating the point of suspension of the blockC. cutting the spring (i.e.changing the length of the spring)D. immersing the mass in a liquid
Answer» Correct Answer - A,C,D Time period of spring mass system `T = 2pisqrt((M)/(K))` Which can be changed (A), (C) & (D)

Discussion

474.	The time period of a particle executing `SHM` is `T`. After a time `T//6` after it passes its mean position, then at`t = 0` its :A. displacement will be one-half of its amplitudeB. velocity will be one-half of its maximum velocityC. kinetic energy `= 1//3` (potential energy)D. acceleration will be `(sqrt(3))/(2)` times of its maximum accleration
Answer» Correct Answer - B,C,D Let `x = Asinomegat` where `omega = (2pi)/(T)` For `(A) = x = A sin ((2pi)/T) (T/6) = Asin((pi)/3) = (sqrt(3))/(2)` For (B) : `v = (dx)/(dt) = Aomegacosomegat` At `t = (T)/(6), v = Aomegacos((2pi)/T) (T/6) = Aomegacos"(pi)/(3) = (A)/(2)` For (C) : `KE = (1)/(2)m ((v_("max"))/(2))^(2) = 1/4(KE)_("max") = 1/4(TE)` & `PE = TE - KE = 3/4 TE rArr KE = 1/3 (PE)` For (D) : Acceleration `a = (dv)/(dt) = - Aomega^(2) sin omegat = - omega^(2) x`

Discussion

475.	Time period of a particle executing `SHM` is `8` sec. At `t=0` it is at the mean position. The ratio of the distance covered by the particle in the `1st` second to the `2nd` second is:A. `1/2`B. `1/sqrt2`C. `sqrt2`D. `(1)/(sqrt2-1)`
Answer» Correct Answer - D `T=8s`, `T/4=25` The particle moves in a straight line without change in direction so displacement and distance are same in first `2s` `t=1s`, `x_1=Asinomegat` `=Asinomega=Asin((2pi)/(T))=Asin((2pi)/(8))=A/sqrt2` `t=2s`, `x_2=A` `x_2-x_1=A(1-sinomega)=A(1-sin((2pi)/(T)))` `A(1-sin((2pi)/(8)))=A(1-(1)/(sqrt2))` `(x_1)/(x_2-x_1)=(A//sqrt2)/(A(1-1/sqrt2))=(1)/(sqrt2-1)`

Discussion

476.	A particle executing simple harmonic motion along y -axis has its motion described by the equation `y = A sin (omega t )+ B`. The amplitude of the simple harmonic motion isA. AB. BC. A+BD. `sqrtA+B`
Answer» Correct Answer - A

Discussion

477.	In order that the resultant path on superimposing two mutually perpendicular SHM be a circle, the conditions are thatA. the amplitudes on both SHM should be equal and they should have a phase difference of `(pi)/(2)`B. the amplitude should be in the ratio 1:2 and the phase difference should be zeroC. the amplitude should be in the ratio 1:2 and the phase difference should be `(pi)/(2)`D. the amplitudes should be equal and the phase difference should be zero
Answer» Correct Answer - A (a) Suppose `x=A "sin" omegat` and `y= A"cos"omegat` Then by squaring and adding these two equations , we get ` x^(2)+y^(2)=A^(2)`

Discussion

478.	Time period of a particle executing `SHM` is `8` sec. At `t=0` it is at the mean position. The ratio of the distance covered by the particle in the `1st` second to the `2nd` second is:A. `(1)/(sqrt(2)+q)`B. `sqrt(2)`C. `(1)/(sqrt(2))`D. `sqrt(2)+1`
Answer» Correct Answer - C (c)`omega=(2pi)/(T)=(pi)/(4)s` . Therefore `y = A " sin"(pi)/(A)t` . Now put x=1 s and then x=2s.

Discussion

479.	A particle executing simple harmonic motion along y -axis has its motion described by the equation `y = A sin (omega t )+ B`. The amplitude of the simple harmonic motion isA. AB. BC. A+BD. `sqrt(A+B)`
Answer» Correct Answer - A (a) The amplitude is a maximum displacement from the mean position

Discussion

480.	x-t equation of a particle in SHM is given as x=`1.0sin(12pit)` in SI units. Potential energy at mean position is zero. Mass of particle is `1/4` kg. Match the following table (SI units).
Answer» Correct Answer - A::B::C::D

Discussion

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

Discussion

482.	Describe the motion of the mass m shown in figure. The walls and the block are elastic.
Answer» The block reaches the sring with a speed v. It now compresses te spring . The block is decelerated due toteh sprng force comes to rest when `1/2mv^2=1/2kx^2` and returns back. It is accelerated due toteh spring force till the spring acquires its N/Atural length. The contact of the block with the spring is now broken. At this instant it has regained its speed v (towards left) as teh sprig is unstretched and no potential energy is stored. This proces takes half the period of oscillation, i.e. `pisqrt(m/k)`. The block strikes the left wall after a time `L/v` and as the collision is elstic it rebounds with the same speed v. After a time `L/v`, it again reaches the spring and the process is repeated. The block thus undergoes periodic motion with time period `pisqrt(m/k)+(2L)/v`

Discussion

483.	A particle, with restoring force proportional to displacement and resulting force proportional to velocity is subjected to a force `F sin omega t`. If the amplitude of the particle is maximum for `omega = omega_(1)`, and the energy of the particle is maximum for `omega=omega_(2)`, thenA. `omega_(1)=omega_(0)and omega_(2) ne omega_(0)`B. `omega_(1)=omega_(0)and omega_(2)=omega_(0)`C. `omega_(1)neomega_(0)and omega_(2)=omega_(0)`D. `omega_(1)neomega_(0)and omega_(2)neomega_(0)`
Answer» Correct Answer - C

Discussion

484.	Given in fig. is the graph between frequency v of the incident light and maximum kinetic energy `(E_(k))` of the emitted photoelectrons. Find the values of (i) threshold frequency and (ii) work function form the graph.
Answer» Correct Answer - A `KE_("max") = hf - phi`

Discussion

485.	Photo electric effect is the phenomenon in whichA. Photons come out of a metal when it is hit by a beam of electronsB. Photons come out of the nucleus of an atom under the action of an electric fieldC. Electrons come out of metal with a constant velocity depending on frequency and intensity of incident lightD. Electrons come out of a metal with different velocity not greater than a certain value which depends only on the frequnecy of the incident light wave and not on its intersity.
Answer» Correct Answer - D In photo electric effect the maximum velocity of `e^(-)` will corresoponding to `KE_("max")` & other are less than it.

Discussion

486.	Which one of the following is true in photoelectric emissionA. Photoelectric current is directly proportiona to the amplitude of light of given frequencyB. Photoelectric current is directly proportional to the intesity of light of a given frequency above threshold valueC. Above the threshold frequency the maximum K.E. of photoelectronsis inversely proportional to the frequency of incident lightD. The threshold frequency depends upon the wavelength of incident light
Answer» Correct Answer - B For threshold frequency, `hv_(0) = phi rArr v_(0) = (phi)/(h)` `rArr KE_("max") = hv - phi = h(v-v_(0))` `I_("saturation") prop n` where `I = nhv`

Discussion

487.	When an electro-magnetic radiation is incident on the surface of metal, maximum kinetic energy of photoelectron depends on-A. Frequency of radiationB. Intensity of radiationC. Both the frequency and intensityD. Polarization of radiation
Answer» Correct Answer - A `KE_("max") = hv - phi`

Discussion

488.	The kinetic energy and the potential energy of a particle executing `SHM` are equal The ratio of its displacement and amplitude will beA. `1/sqrt2`B. sqrt3/2`C. `1/2`D. `sqrt2`
Answer» Correct Answer - A

Discussion

489.	Statement-1 : In compound pendulum, if suspension point and centre of oscillation are mutually inter change, then no change in time period is obtained. Statement-2 : Length of equivalent simple pendulum remains same in both the case.A. Statement-1 is True, Statement-2 is True , Statement-2 is a corrrect explanation for Statement-1B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a corrrect explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A

Discussion

490.	The amplitude and maximum velocity will be respectively `X= 3 sin 2t + 4 cos 2t`The amplitude and maximum velocity will be respectivelyA. 5,10B. 3,2C. 4,2D. 3,4
Answer» Correct Answer - A

Discussion

491.	The displacement equation of a particle is `x=3 sin 2t+4cos2t`. The amplitude and maximum velocity will be respectivelyA. `5,10`B. `3,2`C. `4,2`D. `3,8`
Answer» Correct Answer - A `x = 3sin 2t + 4t + 4cos 2t = 5sin (2t + phi)` `rArr a = 5, v_(max) = aomega = (5)(2) = 10`

Discussion

492.	A body of mass 0.01 kg executes simple harmonic motion (SHM) about x=0 under the influence of a force shown in the figure. The period of the SHM is A. 1.05 sB. 0.52 sC. 0.25 sD. 0.31 s
Answer» Correct Answer - D (d) From graph, slope `K=(F)/(x)=(8)/(2)=4` `T=2pisqrt((m)/(k))impliesT=2pisqrt((0.01)/(4))=0.31s `

Discussion

493.	A body of mass 0.01 kg executes simple harmonic motion about x = 0 under the influence of a force as shown in figure. The time period of SHM is A. 1.05 sB. 0.52 sC. 0.25 sD. 0.30 s
Answer» Correct Answer - D

Discussion

494.	A particle is executing `SHM` of amplitude `A`, about the mean position `X=0`. Which of the following cannot be a possible phase difference between the positions of the particle at `X=+ A//2 and X=-A//sqrt(2)`.A. `75^(@)`B. `165^(@)`C. `135^(@)`D. `195^(@)`
Answer» Correct Answer - C

Discussion

495.	The intensity of X-rays form a Coolidge tube is plotted against wavelength `lambda` as shown in the figure. The minimum wavelength found is `lambda_c` and the wavelength of the `K_(alpha)` line is `lambda_k`. As the accelerating voltage is increased (a) `lambda_k - lambda_c` increases (b) `lambda_k - lambda_c` decreases (c ) `lambda_k` increases (d) `lambda_k` decreasesA. `lambda_(k) - lambda_(c)` increasesB. `lambda_(k) - lambda_(c)` decreasesC. `lambda_(k)` increasesD. `lambda_(k)` decreases
Answer» Correct Answer - A Wavelength `lambda_(k)` is independent of the acceleration voltage (V), while the minimum wavelength `lambda_(c)` is inversely proportional to V. therefore, as V is increased, `lambda_(k)` remains unchanged where as `lambda_(c)` decreases or `lambda_(R)-lambda_(c)` will increase.

Discussion

496.	If the kinetic energy of a free electron doubles , its de - Broglie wavelength changes by the factorA. `1/2`B. `2`C. `1/(sqrt(2))`D. `sqrt(2)`
Answer» Correct Answer - C The de-Brogile wavelengths and kinetic energy of a free electron are related as `lambda=h/sqrt(2mK)implies lambda_(2)/lambda_(1)=sqrt(K_(1)/K_(2))=sqrt(1/2)implies lambda_(2)=lambda_(1)/sqrt(2)`

Discussion

497.	A uniform rod of length l is suspended from a point P. and the rod is made to undergo small oxcillations. Match the following
Answer» Correct Answer - `(A rarr t, B rarr t, C rarr s)` `T=2pisqrt((I)/(mgl))`. Here , l is distance of point P from centre of mass. When P is the centre , l=0. Therefore, `T=oo`, i.e., rod will not oscillate. When P is end point , `I=(ml^(2))/(3)` `therefore T=2pisqrt((l)/(3g))impliesT_("rod")=T_("pendulum")` or `2pisqrt((2l)/(3g))=2pisqrt((L)/(g))` or `L=(2l)/(3)`

Discussion

498.	A uniform rod of length l is suspended from a point P. and the rod is made to undergo small oxcillations. Match the following
Answer» Correct Answer - A::B::C

Discussion

499.	(i) A particle is performing simple harmonic motion with time period T. At an instant its speed is 60% of its maximum value and is increasing. After an interval `Delta t` its speed becomes 80% of its maximum value and is decreasing. Find the smallest value of `Delta t` in terms of T. (ii) A particle is doing SHM of amplitude 0.5 m and period `pi` seconds. When in a position of instantaneous rest, it is given an impulse which imparts a velocity of `1 m//s` towards the equilibrium position. Find the new amplitude of oscillation and find how much less time will it take to arrive at the next position of instantaneous rest as compared to the case if the impulse had not been applied.
Answer» Correct Answer - (i)` Delta T=(t)/(4), (ii) A=(1)/(sqrt(2))m; t=(pi)/(8) s`

Discussion

500.	Two particles move parallel to x - axis about the origin with the same amplitude and frequency. At a certain instant they are found at distance `(A)/(3)` from the origin on opposite sides but their velocities are found to be in the same direction. What is the phase difference between the two ?
Answer» Let equation of two SHM be `x_(1)=A"sin"omegat" "…(i)` `x_(2)=A"sin"(omegat+phi)" "…(ii)` Given, `(A)/(3)=A"sin"omegat` and `-(A)/(3)=A"sin"(omegat+phi)` Which gives `"sin"omegat=(1)/(3)` `"sin"(omegat+phi)=-(1)/(3)` From Eq. (iv), `"sin"omegat" cos"phi+"cos"omegat" sin"phi=-(1)/(3)` `implies (1)/(3)"cos"phi+sqrt(1-(1)/(9))"sin"phi=-(1)/(3)` Solving this equation, we get or `"cos"phi=-1,(7)/(9)` `implies phi=pi` or `"cso"^(-1)((7)/(9))` Differentiating Eqs. (i) and (ii) we obtain `v_(1)=Aomega"cos"omegat` and `v_(2)=Aomega"cos"(omegat+phi)` If we put `phi=pi`, we find `v_(1)` and `v_(2)` are of opposite signs .Hence, `phi=pi` is not acceptance. `therefore phi="cos"^(-1)((7)/(9))`

Discussion

Explore topic-wise InterviewSolutions in .

An block rides on position that is moving vertically with simple harmonic motion. The maximum speed of the piston is 2m/s. At what minimum amplitude of motion will the block and piston separate? (g=10m/`s^(2)`)A. 20 cmB. 30cmC. 40 cmD. 50 cm

A body undergoing `SHM` about the origin has its equation is given by `X=0.2 cos 5pit`. Find its average speed from `t=0 to t=0.7 sec`.

The acceleration-displacement `(a-X)` graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscillation.

The de Broglie waves are associated with moving particles. These particle may beA. electronsB. `He^(+), Li^(2+)` ionsC. Cricket ballD. All of the above

Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)` is close to (Avogadro number `= 6.023 xx 10^(23) // "g. mole"`) :-A. `7.5 xx 10^(21)`B. `1.75 xx 10^(22)`C. `6.2 xx 10^(21)`D. `1.25 xx 10^(22)`

Two particles are executing S.H.M. The equation of their motion are `y_(1) = 10 sin (omega t + (piT)/4),y_(2)=25 sin (omegat+(sqrt3piT/4))` . What is the ratio of their amplitudeA. `1:1`B. `2:5`C. `1:2`D. None of these

The given figure shows the variation of the kinetic energy of a simple pendulum with its angular displacement `(theta)` from the vertical. Mass of the pendulum bob is m = 0.2 kg. Find the time period of the pendulum. Take `g = 10 ms^(-2)`.

If the displacement of simple pendulum at any time is `0.02` m and acceleration is `2m//s^(2)`, then in this time angular velocity will beA. 100 rad/sB. 10 rad/sC. 1 rad/sD. `0.1` rad/s

A simple wave motion represented by `y=5(sin 4pi t+sqrt3 cos 4pit)`. Its amplitude isA. 5B. `5sqrt3`C. `10sqrt3`D. 10

The frequency of a simple pendulum is `n` oscillations per minute while that of another is `(n + 1)` oscillations per minute. The ratio of length of first second is-A. `(n)/(n+1)`B. `(n+(1)/(n))^(2)`C. `((n+1)/(n))^(2)`D. `((n)/(n+1))^(2)`

In a laboratory experiment with simple pendulum it was found it took 36 s to complete 20 oscillations when the effective length was kept that 80 cm. Calculate the acceleration due to gravity from these data.

A simple pendulum has a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of the pendulum willA. Remains unchangedB. IncreaseC. DecreaseD. Become erratic

The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are 2 m/s and` 4 m//s^(2)`. Then angular velocity will beA. 3 rad/secB. 0.5 rad/secC. 1 rad/secD. rad/se

Time period of a spring mass system can be changed byA. changing the massB. acclerating the point of suspension of the blockC. cutting the spring (i.e.changing the length of the spring)D. immersing the mass in a liquid

Time period of a particle executing `SHM` is `8` sec. At `t=0` it is at the mean position. The ratio of the distance covered by the particle in the `1st` second to the `2nd` second is:A. `1/2`B. `1/sqrt2`C. `sqrt2`D. `(1)/(sqrt2-1)`

A particle executing simple harmonic motion along y -axis has its motion described by the equation `y = A sin (omega t )+ B`. The amplitude of the simple harmonic motion isA. AB. BC. A+BD. `sqrtA+B`

Time period of a particle executing `SHM` is `8` sec. At `t=0` it is at the mean position. The ratio of the distance covered by the particle in the `1st` second to the `2nd` second is:A. `(1)/(sqrt(2)+q)`B. `sqrt(2)`C. `(1)/(sqrt(2))`D. `sqrt(2)+1`

A particle executing simple harmonic motion along y -axis has its motion described by the equation `y = A sin (omega t )+ B`. The amplitude of the simple harmonic motion isA. AB. BC. A+BD. `sqrt(A+B)`

x-t equation of a particle in SHM is given as x=`1.0sin(12pit)` in SI units. Potential energy at mean position is zero. Mass of particle is `1/4` kg. Match the following table (SI units).

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

Describe the motion of the mass m shown in figure. The walls and the block are elastic.

Given in fig. is the graph between frequency v of the incident light and maximum kinetic energy `(E_(k))` of the emitted photoelectrons. Find the values of (i) threshold frequency and (ii) work function form the graph.

When an electro-magnetic radiation is incident on the surface of metal, maximum kinetic energy of photoelectron depends on-A. Frequency of radiationB. Intensity of radiationC. Both the frequency and intensityD. Polarization of radiation

The kinetic energy and the potential energy of a particle executing `SHM` are equal The ratio of its displacement and amplitude will beA. `1/sqrt2`B. sqrt3/2`C. `1/2`D. `sqrt2`

The amplitude and maximum velocity will be respectively `X= 3 sin 2t + 4 cos 2t`The amplitude and maximum velocity will be respectivelyA. 5,10B. 3,2C. 4,2D. 3,4

The displacement equation of a particle is `x=3 sin 2t+4cos2t`. The amplitude and maximum velocity will be respectivelyA. `5,10`B. `3,2`C. `4,2`D. `3,8`

A body of mass 0.01 kg executes simple harmonic motion (SHM) about x=0 under the influence of a force shown in the figure. The period of the SHM is A. 1.05 sB. 0.52 sC. 0.25 sD. 0.31 s

A body of mass 0.01 kg executes simple harmonic motion about x = 0 under the influence of a force as shown in figure. The time period of SHM is A. 1.05 sB. 0.52 sC. 0.25 sD. 0.30 s

A particle is executing `SHM` of amplitude `A`, about the mean position `X=0`. Which of the following cannot be a possible phase difference between the positions of the particle at `X=+ A//2 and X=-A//sqrt(2)`.A. `75^(@)`B. `165^(@)`C. `135^(@)`D. `195^(@)`

If the kinetic energy of a free electron doubles , its de - Broglie wavelength changes by the factorA. `1/2`B. `2`C. `1/(sqrt(2))`D. `sqrt(2)`

A uniform rod of length l is suspended from a point P. and the rod is made to undergo small oxcillations. Match the following

A uniform rod of length l is suspended from a point P. and the rod is made to undergo small oxcillations. Match the following

Two particles move parallel to x - axis about the origin with the same amplitude and frequency. At a certain instant they are found at distance `(A)/(3)` from the origin on opposite sides but their velocities are found to be in the same direction. What is the phase difference between the two ?