Q: As shown in the figures, a uniform rod O O^ of length l is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. The springs are conn

aligned & ML ^2 3 + K 2 2 + K =0 \\ & + 15 4 k M =0 \\ & =- ( 15 4 k M ) \\ & _1= 15 ~K 4 M aligned 1 3 ML ^2 +2 ~K ~L 2 ~L 2 =0 aligned & + 3 2 K M =0 \\ & =- 3 2 K M \\ & _2= 3 2 K M \\ & _1 _2 = 15 4 2 3 = 5 2 aligned

Q: Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm . This system is suspended vertically from a rigid ceiling using

Given: m 1 = 30 gm and m 2 = 20 gm . Moment of inertia about the axis of rotation is I = m 1 r 1 2 + m 2 r 2 2 X-coordinate of the centre of mass can be written as, x C M = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0 + 20 × 10 - 3 × 10 × 10 - 2 30 × 10 - 3 + 20 × 10 - 3 = 4 × 10 - 2 m = 4 cm = r 1 Therefore, And r 2 = 10 - 4 cm = 6 cm ∴ I = 30 × 10 - 3 × 16 × 10 - 4 + 20 × 10 - 3

Q: A particle of mass 1 kg is subjected to a force which depends on the position as F → = - k x i ^ + y j ^ kg m s - 2 with k = 1 kg s -

Given here: F → = - k x i ^ + y j ^ kg m s - 2 and m = 1 kg In x-direction, F x = - x = m a x So, acceleration, a x = d 2 x d t 2 = - x Now, for particle executing SHM, displacement along x-direction is ⇒ x = A x sin ω t + ϕ x , here, angular frequency, ω = 1 rad s - 1 and velocity, v x = A x ω cos ω t + ϕ x Given at t = 0 , x = 1 2 m &nbsp

Question 1

A tank contains two immiscible liquids of densities 6 and 2 . The higher density liquid is filled up to a height L/2 from the bottom. A thin rod of density and length L is fully immersed and hinged at

Accepted Answer

Let A be the cross-sectional area of the rod. Mass of the rod, M = A L. Weight of the rod, W = A L g, acting at a distance L/2 from the hinge. When the rod is displaced by a small angle , the restoring torque is provided by the buoyant forces from the two liquids. Buoyant force due to the lower liquid, F_ B1 = (6 ) A (L/2) g = 3 A L g. This force acts at the center of the immersed part in the lower liquid, which is at a distance L/4 from the hinge. Torque due to F_ B1 is _ B1 = F_ B1 (L/4) = (3

Question 2

The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R>r as shown in the figure. The other end of the spring is attached on the periphery o

Accepted Answer

E= 1 2 k(R-r)^2 ^2+m g(R-r)(1- )+ 1 2 m v^2+ 1 2 m r^2 2 ^2 Differentiating wrt t , aligned 0 & = 1 2 k ( R - r )^2 2 ~d dt + mg ( R - r ) d dt (2 ^2 4 )+ 1 2 ~m 2 v dv dt + mr ^2 4 2 ~d dt \ & 0= k ( R - r )^2 ~d dt + mg ( R - r ) d dt + mv dv dt + mr ^2 2 ~d dt aligned Also, d d t = V (R-r) d^2 d t^2 = 1 (R-r) d v d t = 1 R-r a aligned & k ( R - r )^2 ~V R - r + mg ( R - r ) V R - r =- mv r - mr ^2 2 ~V r \ & k ( R - r )+ mg =- 3 2 mr \ & -[ k ( R - r )+ mg ] = 3 2 ~m ( R - r ) d ^2 dt ^2 \

Question 3

As shown in the figures, a uniform rod O O^ of length l is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. The springs are conn

Accepted Answer

aligned & ML ^2 3 + K 2 2 + K =0 \ & + 15 4 k M =0 \ & =- ( 15 4 k M ) \ & _1= 15 ~K 4 M aligned 1 3 ML ^2 +2 ~K ~L 2 ~L 2 =0 aligned & + 3 2 K M =0 \ & =- 3 2 K M \ & _2= 3 2 K M \ & _1 _2 = 15 4 2 3 = 5 2 aligned

Question 4

A block of mass 5 ~kg moves along the x-direction subject to the force F=(-20 x+10) N , with the value of x in metre. At time t=0 ~s , it is at rest at position x=1 ~m . The position and momentum of t

Accepted Answer

F =-20 ( x - 1 2 )=-20 X ( X = x - 1 2 ) Particle will perform SHM about x = 1 2 with =2 rad / sec T = sec . Phase covered in t= 4 second =90^ . Given particle is at rest at x =1 ~m x =1 is extreme position. In 4 sec, it will be at equilibrium x =0.5 ~m and momentum = m A =5 2 0.5=5 ~kg ~m / s Direction will be towards -ve x . Hence option (3)

Question 5

Two point-like objects of masses 20 &#160; gm and 30 &#160; gm are fixed at the two ends of a rigid massless rod of length 10 &#160; cm . This system is suspended vertically from a rigid ceiling using

Accepted Answer

Given: m 1 = 30 &#160; gm  and  m 2 = 20 &#160; gm . Moment of inertia about the axis of rotation is I = m 1 r 1 2 + m 2 r 2 2 X-coordinate of the centre of mass can be written as, x C M = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0 + 20 &#215; 10 - 3 &#215; 10 &#215; 10 - 2 30 &#215; 10 - 3 + 20 &#215; 10 - 3 = 4 &#215; 10 - 2 &#160; m = 4 &#160; cm = r 1 Therefore,  And  r 2 = 10 - 4 &#160; cm = 6 &#160; cm &#8756; &#160; I = 30 &#215; 10 - 3 &#215; 16 &#215; 10 - 4 + 20 &#215; 10 - 3

Question 6

A particle of mass 1 &#160; kg is subjected to a force which depends on the position as  F &#8594; = - k x i ^ + y j ^ &#160; kg &#160; m &#160; s - 2  with  k = 1 &#160; kg &#160; s -

Accepted Answer

Given here:  F &#8594; = - k x i ^ + y j ^ &#160; kg &#160; m &#160; s - 2  and  m = 1 &#160; kg In x-direction,  F x = - x = m a x So, acceleration,  a x = d 2 x d t 2 = - x Now, for particle executing SHM, displacement along x-direction is  &#8658; x = A x sin &#969; t + &#981; x , here, angular frequency,  &#969; = 1 &#160; rad &#160; s - 1  and velocity,  v x = A x &#969; cos &#969; t + &#981; x Given at  t = 0 , &#160; x = 1 2 &#160; m &nbsp

Question 7

On a frictionless horizontal plane, a bob of mass m = 0 . 1 &#160; kg  is attached to a spring with natural length l 0 = 0 . 1 &#160; m . The spring constant is k 1 = 0 . 009 &#160; N &#160;

Accepted Answer

Angular frequency for different values of spring constant will be given by,  &#969; 1 = k 1 m  and  &#969; 2 = k 2 m Time period for a spring of fix spring constant,  k  is given by,  = 2 &#960; m k . As per question, spring constant is having different values, hence we will have to find half of the time period for the given parts, first for the right half part of the oscillations & next for the left half part of the oscillation from the mean position. &#8756; &

Oscillations — Previous Year Questions & Solutions