A particle at the end of a spring executes S.H,M with a period `t_(2)` If the period of oscillation with two spring in .A. `T = t_(1) +t_(2)`B. `T^(2) = t_(1)^(2) +t_(2)^(2)`C. `T^(-1) = t_(1)^(-1) +t_(2)^(-1)`D. `T^(-2) = t_(1)^(-2) +t_(2)^(-2)`

A particle at the end of a spring executes S.H,M with a period `t_(2)`

1.	A particle at the end of a spring executes S.H,M with a period `t_(2)` If the period of oscillation with two spring in .A. `T = t_(1) +t_(2)`B. `T^(2) = t_(1)^(2) +t_(2)^(2)`C. `T^(-1) = t_(1)^(-1) +t_(2)^(-1)`D. `T^(-2) = t_(1)^(-2) +t_(2)^(-2)`
Answer» Correct Answer - B `t_(1) = 2pi sqrt((m)/(k_(1))) and t_(2) 2 pi sqrt((m)/(k_(2)))` in series effective spring constant `k = (k_(1)k_(2))/(k_(1) +k_(2))` `:. Time period T = 2pi sqrt((m(k_(1) + k_(2)))/(k_(1)k_(2)))`…..(i) Now `t_(1)^(2) + t_(2)^(2)= 4pi^(2)m((1)/(k_(1)) + (1)/k_(2)) = (4pi^(2)m(k_(1) + k_(2)))/(k_(1)k_(2))` `t_(1)^(2) + t_(2)^(2)= T^(2)`. [Using equation (ii)]

Discussion

No Comment Found

Related InterviewSolutions

A body is performing simple harmonic motion. Then itsA. average total energy per cycle is equal to its maximum kinetic energy.B. average kinetic energy per cycle is equal to half of its maximum kinetic energy.C. average total energy per cycle is equal to half of its malximum kinetic energy.D. None of these.
A verticle mass-spring system executed simple harmonic ascillation with a period `2s` quantity of this system which exhibits simple harmonic motion with a period of `1sec` areA. velocityB. potential energyC. phese difference between acceleration and displacementD. difference between kinetic energy and potential energy.
Four types of oscillatory system a simple pendlum a physic pendlum a torsional pendlum and a spring mass system each of same time period are taken to the mass if line period will have it unchanged?A. only spring - mass system.B. spring - mass system and torsional pendulum.C. spring - mass system and physical pendulum.D. None of these
A body is executing `SHM` under action of the a force of whose maximum is `50 N`. magnitude of force acting on the particle at the time when its energy is half kineic energy and half potential is (Assume potential energy to be zero at mean position).A. `12.5 sqrt(2) N`B. `12.5 N`C. `25 N`D. `25 sqrt(2) N`
A `4 kg` particle is moving along the x- axis under the action of the force `F = - ((pi^(2))/(16)) x N` At `t = 2 sec ` the particle passes through the origin and `t = 10sec`, the speed is `4sqrt(2)m//s` The amplitude of the motion isA. `(32sqrt(2))/(pi)m`B. `(16)/(pi)m`C. `(4)/(pi)m`D. `(16sqrt(2))/(pi)m`
A particle of mass `10gm` is placed in a potential field given by `V = (50x^(2) + 100)J//kg`. The frequency of oscilltion in `cycle//sec` isA. `(10)/(pi)`B. `(5)/(pi)`C. `(100)/(pi)`D. `(50)/(pi)`
The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in metres. The period of small oscillation (in sec) isA. `2pi`B. `pi`C. pi//2`D. `pi//4`
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. `(5)/(3)`B. `(3)/(5)`C. `(25)/(9)`D. `(16)/(9)`
A mass `M` is suspended from a spring of negiliglible mass the spring is pulled a little and then released so that the mass executes simple harmonic oscillation with a time period `T` If the mass is increases by `m` the time period because `((5)/(4)T)`,The ratio of `(m)/(M)` isA. `9//16`B. `25//16`C. `4//5`D. `5//4`
Two simple pendulum of length `1m` and `16m` respectively are both given small displacement in the same direction of the same instant. They will be phase after one shorter pendulum has complated a oscillations. The value of `n` isA. `1//3`B. `2//3`C. `1`D. `4//3`

A particle at the end of a spring executes S.H,M with a period `t_(2)` If the period of oscillation with two spring in .A. `T = t_(1) +t_(2)`B. `T^(2) = t_(1)^(2) +t_(2)^(2)`C. `T^(-1) = t_(1)^(-1) +t_(2)^(-1)`D. `T^(-2) = t_(1)^(-2) +t_(2)^(-2)`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment