Q: Five identical springs are used in the three configurations as shown in figure. The time periods of vertical oscillations in configurations (a), (b) and (c) are in the ratio.

aligned & T_a=2 m k \\ & T_b=2 m (k / 2) \\ & T_c=2 m 2 k \\ & T_a: T_b: T_c=1: 2 : 1 2 aligned

Q: The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by

aligned & T E=K E+P . E \\ & K E=T E-K E \\ & = 1 2 k^2- 1 2 k^2 \\ & K E= 1 2 m ^2 (A^2-x^2 ) [ k=m ^2 ] aligned KE is maximum at x =0 \& KE is zero when x = A K . E vs x graph is parabola.

Question 1

Five identical springs are used in the three configurations as shown in figure. The time periods of vertical oscillations in configurations (a), (b) and (c) are in the ratio.

Accepted Answer

aligned & T_a=2 m k \ & T_b=2 m (k / 2) \ & T_c=2 m 2 k \ & T_a: T_b: T_c=1: 2 : 1 2 aligned

Question 2

A particle executes simple harmonic motion between x=-A and x=+A. If time taken by particle to go from x=0 to A 2 is 2 s; then time taken by particle in going from x= A 2 to A is:

Accepted Answer

Let time from 0 to A / 2 is t _1 and from A / 2 to A is t _2 From the standard equation of SHM, A 2 = A ( t _1 ) t_1= ^ -1 ( 1 2 )= 6 ....(i) Using x=A_0 t again A=A (t_1+t_2 ) (t_1+t_2 )= ^ -1 (1)= 2 Using (i) t_2= 2 - 6 = 3 .....(ii) Dividing equation (i) by (ii) we get t_1 t_2 = 1 2 t_2=2 t_1=2 2=4 sec

Question 3

A simple pendulum doing small oscillations at a place R height above earth surface has time period of T_1=4 ~s . T_2 would be it's time period if it is brought to a point which is at a height 2 R from

Accepted Answer

Time period of simple pendulum, T =2 l ~g \& ~g = GM ( R + h )^2 T ( R + h ) T _1 ~T _2 = ( R + h _1 ) ( R + h _2 ) = ( R + R ) ( R +2 R ) = 2 3 3 ~T _1=2 ~T _2

Question 4

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by

Accepted Answer

aligned & T E=K E+P . E \ & K E=T E-K E \ & = 1 2 k^2- 1 2 k^2 \ & K E= 1 2 m ^2 (A^2-x^2 ) [ k=m ^2 ] aligned KE is maximum at x =0 \& KE is zero when x = A K . E vs x graph is parabola.

Question 5

A particle executes SHM of amplitude A. The distance from the mean position when it's kinetic energy becomes equal to its potential energy is

Accepted Answer

Let the distance from the mean position is X . Given KE = PE So, 1 2 M ^2 (A^2-x^2 )= 1 2 M ^2 x^2 A^2-x^2=x^2 A^2=2 2 x = A 2

Question 6

In a resonance column first and second resonance are obtained at depths 24 ~cm and 78 ~cm the third resonance will be obtained at depth.

Accepted Answer

Given, first resonance column at depth, l_1=24 ~cm Second resonance column at depth, l_2=78 ~cm Third resonance column at depths, l_3= ? As aligned & l_1+x= 4 =24 (i)\ & l_2+x= 3 4 =78 (ii)\ & l_3+x= 5 4 (iii) aligned From Eqs. (i) and (ii), x= l_2-3 l_1 2 = 78-3(24) 2 =3 (iv) From Eqs. (iii) and (i), l_3+x l_1+x =5 l_3=5 l_1+5 x-x=5 l_1+4 x Substituting l_1 and x value, we get l_3=5(24)+4(3) l_3=132 ~cm .

Question 7

Four massless springs whose force constants are 2 k, 2 k, k and 2 k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizonta

Accepted Answer

The equivalent system is shown in figure. The equivalent force constant of which k_ e =k+3 k=4 k = k_ e m = 4 k m =2 k m and f = 2 = 2 2 k m = 1 k m

Question 8

In a simple pendulum of length I the bob is pulled aside from its equilibrium position through an angle and then released. The bob passes through the equilibrium position with speed

Accepted Answer

If l is length of pendulum and be angular amplitude then from principle of conservation of energy (Initial total energy) 0+ mgh = 1 2 m v^ 2 +0 (Final total energy) v= 2 g h = 2 g l(1- )

Oscillations — Previous Year Questions & Solutions