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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Assertion : Average kinetic energy in one oscillation during SHM of a body is `(1)/(4)momega^(2)A^(2)`. Reason : Maximum kinetic energy is `(1)/(2)momega^(2)A^(2)`.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B (b) If `v=omegaA " sin"omegat` Then, `v_("max")=omegaA` `therefore K_("max")=(1)/(2)mv_("max")^(2)=(1)/(2)momega^(2)A^(2)` Further , `gtAlt = lt (1)/(2)momega^(2)A^(2)"sin"^(2)omegat gt ` `(1)/(2)momega^(2)A^(2) lt "sin"^(2)omegatlt ` But `lt "sin"^(2) gt =(1)/(2)` in one oscillation `therefore lt K gt =(1)/(4)momega^(2)A^(2)` |
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| 152. |
A particle performing SHM is found at its equilibrium position at t = 1s and it is found to have a speed 0.25 = m//s at t=2s. If the period of oscillation is 6. Calculate amplitude of oscillation.A. `3/(2pi)m`B. `3/(4pi)m`C. `6/pi`mD. `3/(8pi)`m |
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Answer» Correct Answer - A |
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| 153. |
`A` mass `M` is suspended from a spring of negligible mass. The spring is pulled a little then released, so that the mass executes simple harmonic motion of time period `T`. If the mass is increased by `m`, the time period becomes `(5T)/(3)`. Find the ratio of `m//M`. |
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Answer» Correct Answer - A `T = 2pi sqrt((M)/(k))` `(5T)/(3) = 2pi sqrt ((M + m)/(k))` Solving these two equation, we get `(m)/(M) = (16)/(9)` |
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| 154. |
A mass (M) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes `(5T)/3`. Then the ratio of `m/M` is .A. `3/5`B. `25/9`C. `16/9`D. `5/3` |
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Answer» Correct Answer - C `T = 2pisqrt((M)/(k))` and `(5T)/(3) = 2pisqrt((M + m)/(k))` On driving ` = (5T)/(3T) = sqrt((M + m)/(M))` `rArr (25)/(9) = (M + m)/(M) rArr (m)/(M) = (16)/(9)` |
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| 155. |
Two pendulum of lengths `1m` and `16m` are in phase at the mean position at a certain instant of time. If T is the time period of the shorter pendulum, then the minimum time after which they will again be in phase isA. `T/3`B. `(2T)/(3)`C. `(4T)/(3)`D. `(8T)/(3)` |
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Answer» Correct Answer - C `T_1=2pisqrt(1/g)=T`, `T_2=2pisqrt((16)/(g))=4T` `(omega_1-omega_2)t=2pi` `((2pi)/(T)-(2pi)/(4T))t=2pi` `t=(4T)/(3)` |
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| 156. |
(a) A particle executes simple harmonnic motion with an amplitude of `10cm`. At what distance from the mean position are the kinetic and potential energies equal? (b) The maximum speed and acceleration of a particle executing simple harmonic motion are `10m//s` and `50cm//s`. Find the position(s) of the particle when the speed is `8cm//s`. |
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Answer» (a) `K=U` `1/2momega^2(A^2-x^2)=1/2momega^2x^2` `A^2-x^2=x^2` `x^2=A^2/2` `x=+-A/sqrt2=+-(10)/(sqrt2)=+-5sqrt2cm` (b) `v_(max)=Aomega=10` (i) `a_(max)=Aomega^2=50` (ii) `(ii)//(i)impliesomega=5rad//s, A=2cm` `v=omegasqrt(A^2-x^2)` `8=5sqrt((2)^2-x^2)` `x^2=(2)^2-(1.6)^2=1.44` `x=+-1.2cm` |
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| 157. |
Starting from the origin a body osillates simple harmonicall with a period of 2 s. A fter what time will its kinetic energy be 75% of the total energy?A. `1/4s`B. `1/3s`C. `(1)/(12)s`D. `1/6s` |
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Answer» Correct Answer - D `K=3/4E` `1/2momega^2(A^2-x^2)=3/4*1/2momega^2A^2impliesx=+-A/2` `x=Asin((2pi)/(T))timpliesA/2sin((2pi)/(2))t=Asinpit` `sin pi t=1//2=sinpi//6impliest=1//6s` |
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| 158. |
Starting from the origin a body osillates simple harmonicall with a period of 2 s. A fter what time will its kinetic energy be 75% of the total energy?A. (a) `1/6s`B. (b) `1/4s`C. (c )`1/3s`D. (d) `1/(12)s` |
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Answer» Correct Answer - A (a) K.E of a body undergoing SHM is given by, K.E.=1/2ma^(2)omega^(2) cos^(2) omegat, T.E.=1/2ma^(2)omega^(2)` Given K.E.=0.75T.E`. `rArr0.75=cos^(2)omegat rArromegat=(pi)/6` `rArrt=(pi)/(6xxomega)rArr t=(pixx2)/(6xx2pi)rArr t=1/6s` |
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| 159. |
A simple harmonic motion is given by `y = 5(sin3pit + sqrt(3) cos3pit)`. What is the amplitude of motion if `y` is in `m` ?A. `100 cm`B. `5 cm`C. `200 cm`D. `1000 cm` |
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Answer» Correct Answer - D `y = 10 ((1)/(2)sin3pit + sqrt(3)/(2)cos3pit) = 10 sin (3pit + (pi)/(3))` thus amplitude is `10 m` or `1000 cm` |
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| 160. |
Starting from the origin a body osillates simple harmonicall with a period of 2 s. A fter what time will its kinetic energy be 75% of the total energy?A. `(1)/(12)s`B. `(1)/(6)s`C. `(1)/(4)s`D. `(1)/(3)s` |
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Answer» Correct Answer - B |
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| 161. |
Starting from the origin a body osillates simple harmonicall with a period of 2 s. A fter what time will its kinetic energy be 75% of the total energy?A. `1/6s`B. `1/4s`C. `1/3s`D. `1/12s` |
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Answer» Correct Answer - A `KE = (75/100)TE` `rArr 1/2momega^(2)(A^(2) - x^(2)) = 3/4[1/2momega^(2)A^(2)] rArr x = (A)/(2)` Now from `x = Asinomegat` we have `A/2 = Asin(omegat) rArr ((2pi)/(T)) t = (pi)/(6) rArr t = 1/6s` |
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| 162. |
The position of a particle in motion is given by `y = Csinomegat + Dcosomegat w.r.t` origin. Then motion of the particle isA. `SHM` with amplitude `C+D`B. `SHM` with amplitude `sqrt(C^(2) + D^(2))`C. `SHM` with amplitude `((C + D))/(2)`D. not `SHM` |
| Answer» Correct Answer - B | |
| 163. |
When two mutually perpendicular simple harmonic motions of same frequency, amplitude and phase are susperimposedA. the resulting motion is uniform circular motion.B. the resulting motion is a linear simple harmonic motions along a straight ine inclined equally to the straight lines of motion of component ones.C. the resulting motion is an elliptical motion, susmmetical about the lines of motion of the compounents.D. the two `S.H.M.` will cancel each other. |
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Answer» Correct Answer - B `y = A sin(omegat + phi)` and `x = A sin (omegat + phi)` then `y = x` so path is straight line. |
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| 164. |
Two simple harmonic motions `y_(1) = Asinomegat` and `y_(2)` = Acos`omega`t are superimposed on a particle of mass m. The total mechanical energy of the particle isA. `1/2momega^(2)A^(2)`B. `momega^(2)A^(2)`C. `1/4momega^(2)A^(2)`D. zero |
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Answer» Correct Answer - B |
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| 165. |
A particle is subjected to two mutually perpendicular simple harmonic motions such that its `X` and `y` coordinates are given by `X=2 sin omegat` , `y=2 sin (omega+(pi)/(4))` The path of the particle will be:A. an ellipseB. a straight lineC. a parabolaD. a circle |
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Answer» Correct Answer - A |
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| 166. |
When two mutually perpendicular simple harmonic motions of same frequency, amplitude and phase are superimposed.A. the resulting motion is uniform circular motion.B. the resulting motion is a linear simple harmonic motions along a straight ine inclined equally to the straight lines of motion of component ones.C. the resulting motion is an elliptical motion, susmmetical about the lines of motion of the compounents.D. the two `S.H.M.` will cancel each other. |
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Answer» Correct Answer - B `y = A sin(omegat + phi)` and `x = A sin (omegat + phi)` then `y = x` so path is straight line. |
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| 167. |
A particle is placed at the lowest point of a smooth wire frame in the shape of a parabola, lying in the vertical xy-plane having equatioin `x^(2)`=5y(x,y are in meter). After slight displacement, the particle is set free. Find angular frequency of osciallation (in rad/sec) (Take g=10 m/`s^(2))`A. 2 rad/sB. 4 rad/sC. 6 rad/sD. 8 rad/s |
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Answer» Correct Answer - A |
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| 168. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. Proportional to `1/sqrt(a)`B. inderpendent of aC. proportional to `sqrta`D. proportional to `a^(3//12)` |
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Answer» Correct Answer - A |
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| 169. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proprtional to `1//sqrta`B. independent of aC. proportional to `sqrta`D. proportional to `a^(3//2)` |
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Answer» Correct Answer - A |
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| 170. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proportional to `(1)/(sqrta)`B. independent of aC. proportional to `sqrta`D. proportaional to `a^(3//2)` |
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Answer» Correct Answer - A `u=k|x|^(3)` `U_("max")=ka^(3)` The oscillation energy is given by `(1)/(2)ma^(2)omega^(2)`. `therefore" "(1)/(2)ma^(2)omega^(2)=ka^(2)a` or `omega^(2)=(2k)/(m)a` or `omega=sqrt([((2k)/(m))a])` `therefore" "T=(2pi)/(omega)=2pisqrt((m)/(2ka))` `therefore" "Tprop(1)/(sqrta)` |
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| 171. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. porprotional to `(1)/(sqrt(a))`B. independent of aC. proportional to `sqrt(a)`D. proportional to `a^(3//2)` |
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Answer» Correct Answer - A `U(x) = k|x|^(3)` `:. [k] = (|U|)/(|x^(3)|) = ([ML^(2)T^(-2)])/([L^(3)]) = [ML^(-1)T^(-2)]` Now, time period may depend on `T prop ("mass")^(x)("amplitude")^(y) (k)^(z)` `rArr [M^(0)L^(0)T] = [M]^(x)[L]^(y)[ML^(-1)T^(-2)]^(2) = [M^(x+2)L^(y-2)T^(-2x)]` Equating powers, we get `2z = 1` or `Z = -1//2` `y -z = 0` or `y = z = -1//2` Hence, `Tprop("amplitude")^(-1//2)` `rArr T prop (a)^(-1//2) rArr T prop (1)/(sqrt(a))` |
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| 172. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proportional to `1sqrt(a)`B. independent of `a`C. proportional to `sqrt(a)`D. proportional to `a^(3//2)` |
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Answer» `V=k|x|^(3)` `F=(dv)/(dx)=-3k|x|^(2)`…….`(1)` The equation of simple harmonic motion is given as `x=a sinomega t` `rArrm(d^(2)x)/(dt^(2))=m(-aomega^(2)sinomegat)=-momega^(2).x`……`(2)` Using `(1)` and `(2)`, we obtain `3k|x|^(2)=momega^(2)xrArromega=sqrt(3kx//m)` `rArr T=2pisqrt((m)/(3kx))rArrT=2pisqrt((m)/(3kasinomegat))` `T prop (1)/(sqrt(a))` |
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| 173. |
Radioactive nuclei emit `beta^-1` particles. Electrons exist inside the nucleus.A. Statement-1 is True, Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - C | |
| 174. |
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)` cmB. `sqrt(3)` cmC. 1 cmD. 2 cm |
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Answer» Correct Answer - D (d) At mean position , velocity is maximum `v_("max") =omegaaimpliesomega=(v_("max"))/(a)=(16)/(4)=4` `therefore v=sqrt(a^(2)-y^(2))implies 8sqrt(3)=4sqrt(4^(2)-y^(2))` `192=16(16-y^(2))implies12=16-y^(2)impliesy=2 cm` |
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| 175. |
A simple pendulum has some time period `T`. What will be the percentage change in its time period if its amplitude is decreased by `5%`A. `6 %`B. `3 %`C. `1.5 %`D. `0 %` |
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Answer» Correct Answer - D `T = 2pisqrt((l)/(g))`, As it does not depend on amplitude `:. %` change in time period is `0%` Hence option `(D)` is correct. |
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| 176. |
Assertion : The period change in time period is `1.5% ` if the length of simple pendulum increases by `3%`. Reason : Time period is dinesty proportional to length of pendulum.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - C |
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| 177. |
Passage XIV) A uniform cylindrical block of mass 2M and cross-sectional area A remains partially submerged in a non viscous liquid of density `rho`, density of the material of the cylinder is `3rho`. The cylinder is connected to lower end of the tank by means of a light spring of spring constant K. The other end of the cylinder is connected to anotehr block of mass M by means of a light inextensible sting as shown in the figure. The pulleys shown are massless and frictionless and assume that the cross-section of the cylinder is very small in comparison to that of the tank. Under equilibrium conditions, half of the cylinder is submerged. [given that cylinder always remains partially immersed) Under equilibrium conditionsA. extensions of the spring is `(Mg)/(3K)`B. compression of the spring is `(2Mg)/(3K)`C. compression of the spring is `(Mg)/8k)`D. extension of the spring is `(Mg)/(6k)` |
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Answer» Correct Answer - B |
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| 178. |
Passage XIV) A uniform cylindrical block of mass 2M and cross-sectional area A remains partially submerged in a non viscous liquid of density `rho`, density of the material of the cylinder is `3rho`. The cylinder is connected to lower end of the tank by means of a light spring of spring constant K. The other end of the cylinder is connected to anotehr block of mass M by means of a light inextensible sting as shown in the figure. The pulleys shown are massless and frictionless and assume that the cross-section of the cylinder is very small in comparison to that of the tank. Under equilibrium conditions, half of the cylinder is submerged. [given that cylinder always remains partially immersed) By what maximum distance cylinder will be pushed downward into the liquid from equilibrium position so that when it is set free then tension in the string will not vanish [Assume at equilibrium position system was at rest]A. `(3Mg)/(K+Arhog)`B. `(3Mg)/(2(K+Arhog))`C. `(8Mg)/(3(K+Arhog))`D. `(3Mg)/(2K + 3Arhog)` |
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Answer» Correct Answer - A |
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| 179. |
Passage XIV) A uniform cylindrical block of mass 2M and cross-sectional area A remains partially submerged in a non viscous liquid of density `rho`, density of the material of the cylinder is `3rho`. The cylinder is connected to lower end of the tank by means of a light spring of spring constant K. The other end of the cylinder is connected to anotehr block of mass M by means of a light inextensible sting as shown in the figure. The pulleys shown are massless and frictionless and assume that the cross-section of the cylinder is very small in comparison to that of the tank. Under equilibrium conditions, half of the cylinder is submerged. [given that cylinder always remains partially immersed) If the cylinder is pushed down from equilibrium by a distance which is half the distance as calculated in the above question, determine time period of subsequent motion.A. `(2pisqrt((3M)/(2(K+Arhog)B. `(2pisqrt(M/(2K+Arhog))`C. `2pisqrt((3M)/(K+Arhog))`D. `2pisqrt((2M)/(K+3Arhog))` |
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Answer» Correct Answer - C |
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| 180. |
If x, and a denote the displacement, the velocity and the acceler of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?A. (a) `aT//x`B. (b) `aT+2piv`C. (c ) `aT//v`D. (d) a^(2)T^(2)+4pi^(2)v^(2)` |
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Answer» Correct Answer - A (a) For an SHM, the acceleration `a=-omega^(2)x` where `omega^(2)` is a constant. Therefore `a/x` is a constant. The time period (T) is also constant. Therfore `(aT)/x` is a constant. |
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| 181. |
If x, and a denote the displacement, the velocity and the acceler of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?A. `aT//X`B. `aT+2pi V`C. `aT//v`D. `a^(2)T^(2)+4pi^(2)V^(2)` |
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Answer» Correct Answer - C |
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| 182. |
If x, and a denote the displacement, the velocity and the acceler of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?A. `a^(2)T^(2)+4pi^(2)V^(2)`B. `aT//X`C. `aT+2piV`D. `aT//V` |
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Answer» Correct Answer - B |
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| 183. |
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?A. `aT + 2piv`B. `(aT)/(v)`C. `a^(2)T^(2)+4pi^(2)v^(2)`D. `(aT)/(x)` |
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Answer» Correct Answer - C::D `a = -omega^(2)x, v = omegasqrt(A^(2)-x^(2))` `a^(2)T^(2) + 4pi^(2)v^(2) = omega^(2)x^(2)T^(2) + 4pi^(2)omega^(2)(A^(2) - x^(2))` `= (4pi^(2))/(T^(2)) omega^(2)x^(2)T^(2)+ 4pi^(2)omega^(2)(A^(2) - x^(2))` `= 4pi^(2)omega^(2)A^(2) = "constant"` `(aT)/(x) = -(omega^(2)xT)/(x) = -omega^(2)T = "constant"` |
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| 184. |
The time period of a simple pendulum of length 9.8 m isA. 0.159 sB. 3.14 sC. 6.5 sD. 6.28 s |
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Answer» Correct Answer - D (d) Time period `T=2pisqrt((k)/(m))` Given , `g=9.8 ms^(-2), l=9.8 m` `T=2pisqrt((9.8)/(9.8))=2pi=2xx3.14=6.28 s` |
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| 185. |
A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude A. What will be the frequency of oscillation, the block will just start to slip? Coefficient of friction`=mu`.A. `(1)/(2pi)sqrt((mug)/(A))`B. `(1)/(4pi)sqrt((mug)/(A))`C. `2pisqrt((A)/(mug))`D. `4pisqrt((A)/(mug))` |
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Answer» Correct Answer - A (a) When restoring force will become equal to the frictional force, block will start to slip. `therefore` Restoring force=Friction force `implies kA=mumg" "[because F_("rest"=-kx]...(i)` Now, frequency , `f=(1)/(2pi)sqrt((k)/(m))` From Eq. , we get `f=(1)/(2pi)sqrt((mug)/(A))` |
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| 186. |
A particle is subjecte to two simple harmonic motions gilven by `x_1=2.0sin(100pit_)and x_2=2.0sin(120pi+pi/3)`, where x is in centimeter and t in second. Find the displacement of the particle at a. t=0.0125, b. t= 0.025. |
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Answer» Correct Answer - A::B::C::D `x_1=2sin100pit` `x_2=2sin(120pit+pi/3)` so, resultant displacement is given by `x=x_1+x_2` `=[sin(100pit)+sin(120pit+pi/3)]` a. At t=0.0125s `x=2[(sin(100pix0.0125+sin(120ixx0.0125+pi/3))]` `=2[sin(5pi)/4+sin((3pi)/2)+pi/3]` `=2[(-0.707)+(-0.5)]` `=2xx(-1.207)=-2.41cm` b. At t=0.025s `x=2[(sin(100pixx0.025)+sin(120pixx0.025+pi/3)]` `=2[sin((10pi)/4)+sin(3pi+pi/3)]` `=2[1+(-0.866)]` `=2xx(0.134)=0.27cm` |
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| 187. |
A particle is subjected to two simple harmonic motions `x_1=A_1 sinomegat` `and x_2=A_2sin(omegat+pi/3)` Find a the displacement at t=0, b. the maxmum speed of the particle and c. the maximum acceleration of the particle |
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Answer» a. At t=0, `x_1=A_1sinomegat=0` `and x_2=A_2sin(omegat+pi/3)` `=A+2sin(pi/3)=(A_2/sqrt3)/2` Thus the resultant displacement at t=0 is `x=x_1+x_2=(A_2sqrt3)/2` b. The resultant of the two motions is a simple harmonic motion of the same angular frequency omega. The amplitude of the resultant motion is `A=sqrt(A_1^2+A_2^2+2A_1A_2cos(pi/3))` `=sqrt(A_1^2+A_2^2+A_1A_2)` the maximum speed is `=v_(max)=Aomega=omegasqrt(A_1^2+A_2^2+A_1A_2)` c. The maximum acceleration is `a_(ma)=Aomega^2=omega^2 sqrt(A_1^2+A_2^2+A_1A_2)` |
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| 188. |
A particle moves along the `X`-axis according to to the law `S=a sin^(2)(omegat-pi//4)` The time period of oscillations isA. `(2pi)/(w)`B. `T//W`C. `(3pi)/(2W)`D. `(pi)/(2W)` |
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Answer» Correct Answer - B |
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| 189. |
Assertion : In x-=3+4 `"cos" omegat` , amplitude of oscillation is 4 units. Reason : Mean position is at x=3.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B (b) `X=x-4=4"cos"omegatimpliesX=0 " at " x=3` |
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| 190. |
A particle moves along the `X`-axis according to to the law `S=a sin^(2)(omegat-pi//4)` The amplitude of the oscillation isA. `a`B. `(a)/(2)`C. `(3a)/(2)`D. `2a` |
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Answer» Correct Answer - B |
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| 191. |
The oscillation of a body on a smooth horizental surface is respresented by the equation `X = A cos (omegat)` where one of the following graph shown correctly the variation a with `t`?A. B. C. D. |
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Answer» Correct Answer - C (c)n`a=-omega^(2)x=-omega^(2)A"cos"omegat` `therefore` a-t graph is -`"cos"omegat` graph |
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| 192. |
The disk has a weight of `100 N`and rolls without slipping on the horizontal surface as it oscillates about its equilibrium position. If the disk is displaced, by rolling it counterclockwise `0.4rad`, determine the equation which describes its oscillatory motion when it is released. |
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Answer» Correct Answer - A::C::D In the displaced position, ` E = (1)/(2)mv^(2) + (1)/(2) Iomega ^(2) + (1)/(2)k(2x)^(2)` `I = (1)/(2) mR^(2)` and `omega = (v)/(R)` `:. E = (3)/(4)mv^(2) + 2kx^(2)` `E =` constant `:. (dE)/(dt) = 0` or `(3)/(2)mv (dv)/(dt) + 4kx (dx)/(dt) = 0` Substituting, `(dx)/(dt) = v` and `(dv)/(dt) = a` `a = - (8k)/(3m).x` Comparing with, `a = - omega^(2)x` We have `omega = sqrt ((8k)/(3m)) = sqrt(((8 xx 1000)/(3 xx 100))/(9.8)) = 16.16rad//s` `:. theta = theta_(0)cos omega t` or `theta = 0.4cos (16.16t)`. |
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| 193. |
On a planet a freely falling body takes 2 sec when it is dropped from a height of 8 m , the time period of simple pendulum of length 1 m on that planet isA. 3.14 secB. 16.28 secC. 1.57 secD. none of these |
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Answer» Correct Answer - A |
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| 194. |
A simple pendulum executing S.H.M. is falling freely along with the support. ThenA. Its periodic time decreasesB. Its periodic time increasesC. It does not oscillate at allD. None of these |
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Answer» Correct Answer - C |
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| 195. |
A simple pendulum executing S.H.M. is falling freely along with the support. ThenA. it does not oscillate at allB. its periodic time increaseC. its periodic time decreaseD. None of the above |
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Answer» Correct Answer - A (a) Time period of a simple pendulum, `T=2pisqrt((l)/(g))` For freely falling effective , g=0 So, `T=oo` `therefore` Pendulum does not oscillate at all. |
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| 196. |
The amplitude of a damped oscillator decreases to `0/9` times ist oringinal magnitude in `5s`, In anothet `10s` it will decrease to a times its original magnitude, wherea eqcrease .A. `0.729`B. `0.6`C. `0.7`D. `0.81` |
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Answer» Correct Answer - A |
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| 197. |
A linear harmonic oscillator of force constant `2 xx 10^6 N//m` and amplitude (0.01 m) has a total mechanical energy of (160 J). Its.A. maximum potential energy is (100 J)B. maximum kinetic energy is (100 J)C. maximum potential energy is (160 J)D. maximum potential energy is zero. |
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Answer» Correct Answer - B::C The total energy of the oscillator `=(1)/(2) kA^2 = Max.K.E.` `= (1)/(2) xx 2 xx 10^6 xx (0.01)^2 = 100 J` As total mechanical energy = (160 J) The (P.E.) at equilibrium position is not zero. (P.E.) at mean position = `(160 - 100) J = 60 J` :. Max (P.E.) = `(100 + 60) J = 160 J`. |
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| 198. |
A linear harmonic oscillator of force constant `2 xx 10^6 N//m` and amplitude (0.01 m) has a total mechanical energy of (160 J). Its.A. Maximum potential energy is 100 JB. Maximum K.E. is 100 JC. Maximum P.E. is 40 JD. Minimum P.E. is zero |
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Answer» Correct Answer - B Maximum kinetic energy =`1/2 Ka^(2)=100 J` |
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| 199. |
A linear harmonic oscillator of force constant `2 xx 10^6 N//m` and amplitude (0.01 m) has a total mechanical energy of (160 J). Its.A. Maximum potential energy is 100 JB. maximum kinetic energy is 100 JC. maximum potential energy is 160 JD. maximum potential energy is zero |
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Answer» Correct Answer - B::C |
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| 200. |
Write the equation of SHM for the situations show below: |
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Answer» (a) At `t = 0, x = +A` `x = Asin(omegat + phi)` `A = Asin(phi)` `phi = pi//2` `x = A sin (omegat + (pi)/(2)) = Acos(omegat)` (b) At `t = 0, x = -A` `x = A sin(omegat + phi)` `-A = Asin phi` `phi = (3pi)/(2)` `x = A sin(omegat + (3pi)/(2))` `x = -A cos (omegat)` (c) At `t = 0, x = (A)/(2)` `x = A sin (omegat + phi)` `(A)/(2) = A sin (omegat + phi)` `(1)/(2) = sin phi rArr phi = 30^(@), 150` Particle is moving towards the mean position and in negative direction. velocity `v = Aomega cos (omegat + phi)` At `t = 0, v = -ve` hence `v = Aomegacis phi` hence `phi = 150^(@)` `x = A sin(omegat + 150^(@))` (a) `x = Acosoemgat`, (b) `x = -A cosomegat`, (c) `x = Asin(omegat + 150^(@))` |
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