Let us discuss some typical multiple choice questions on oscillations and simple harmonic motion.

(1) A spring of negligible mass and force constant *k* is suspended from a rigid support and a mass *m* is attached to its free end. On depressing the mass slightly and releasing, the system executes simple harmonic oscillations of frequency *f*. The spring is now cut into two pieces with lengths in the ratio 1:2. If the mass *m* is attached to the longer piece, the frequency of oscillations will be

(a) 2*f*

(b) (2/3) *f*

(c) (3/2) *f*

(d) √(3/2) *f*

(e) **√**(2/3) *f*

The period of oscillation of a spring mass system is given by

*T* = 2π√(*m*/*k*)

where *k *is the force constant of the spring. The frequency of oscillations is therefore given by

*f* = (1/2π) √(*k*/*m*)

When the spring is cut in the ratio (of lengths) 1:2, the force constants of the pieces are 3*k* and (3/2)*k*.

[You should note that the spring constant is inversely proportional to the length of the spring. Therefore, when the length of the spring becomes one-third, the spring constant becomes three times (*3k*)* *and when the length becomes two-thirds, the spring constant becomes (3/2)*k*].

The frequency therefore becomes (1/2π) √(3*k */2*m*) = **√**(3/2) *f*.

** ****(2)** One end of a spring of force constant *k* is fixed to a rigid support S on an inclined plane of angle *θ* and the other end is connected to a block of mass *m* as shown in the adjacent figure. The inclined plane is smooth. If the block is displaced along the plane through a small distance and released, the mass executes simple harmonic oscillations. The period of oscillations is

(a) 2π√(*m*/*k*)

(b) 2π√(*m*/*k *sin*θ*)

(c) 2π√(*m*/*k *cos*θ*)

(d) 2π√(*m *sin*θ*/*k*)

(e) 2π√(*m*g sin*θ */*k*)

The period of oscillation of the spring mass system is independent of the gravitational field since the restoring force is supplied by the elastic force in the spring. The period is therefore independent of the direction of motion and is given by

*T* = 2π√(*m*/*k*)* *

** **

(3) A wooden cube of side *a *and mass *m* floats in a liquid of density *ρ*. On giving a small vertical displacement from the equilibrium position, the cube oscillates with a period *T*. Then* T* is

(a) directly proportional to *aρ*

(b) directly proportional to √(*aρ*)

(c) inversely proportional to √(*aρ*)

(d) inversely proportional to *a*√*ρ*

(e) independent of *a*

The period of oscillation is given by

*T* = 2π√(*m*/*k*)* *where *m* is the mass of the cube and k is the force per unit displacement of the block.

When you push the floating cube down through a small distance *x*, the additional force of buoyancy on the cube is equal in magnitude to the weight of the additional liquid displaced by the cube. This is *vρg* where *v* = extra volume of liquid displaced = *a*^{2}x.

Thus, the restoring force on the cube = *vρg =* *a*^{2}xρg. * *

Therefore, the force per unit displacement of the cube, *k * = *a*^{2}xρg/*x* = *a*^{2}ρg

The period of oscillation is thus given by *T* = 2π√(*m*/*k*) = 2π√(*m*/*a*^{2}ρg)

The period is therefore inversely proportional to* a*√*ρ* [Option (d)]

[Note that the period of oscillation is 2π√(*m*/*Aρg*) where *A* is the area of cross section. This expression holds in the case of any block of uniform cross section area *A*].

(4) A simple pendulum is arranged inside an elevator. When the elevator moves upwards with a uniform velocity of 1 ms^{–1}, the period of oscillation is* T*. If the elevator moves upwards with an *acceleration* of 1 ms^{–2}, the period of oscillation of the pendulum will be (assuming *g *= 10 ms^{–2})

(a) *T* ×√(10/11)

(b) *T* ×√(11/10)

(c) *T* ×√11)

(d) *T* ×√10

(e)** ***T *

The period of oscillation of simple pendulum is given by

*T* = 2π√(*ℓ*/*g*) where *ℓ* is the length of the pendulum and *g* is the acceleration due to gravity. The period of the pendulum is given by the above equation when the elevator is at rest and also when it is moving with *uniform velocity*.

Since *g* = 10 ms^{–2}, we have *T* = 2π√(*ℓ*/10).

When the elevator moves up with acceleration (*a*) of 1 ms^{–2}, the weight of the bob of mass *m *increases from *mg* to *m*(*g*+*a*) so that the restoring force on the bob increases correspondingly. The period of oscillation therefore decreases.

In place of *g* the value (*g*+*a*) is to be substituted in the expression for period.

The modified period (*T*_{1})* *is given by

*T*_{1} = 2π√[*ℓ*/(10+1)] = 2π√(*ℓ*/11) = 2π√[(*ℓ*/10) ×(10/11)] =* T* ×√(10/11).

(5) The displacement (*y*) of a particle is given by the equation,

*y* = 5 (cos^{2 }3πt – sin^{2 }3πt). The motion of the particle is

(a) not simple harmonic

(b) simple harmonic with frequency 1 Hz

(c) simple harmonic with frequency 3 Hz

(d) simple harmonic with frequency 6 Hz

(e)** **simple harmonic with frequency 3π Hz

If you are a little bit strong in trigonometry, you will get the correct option (c) easily. We have cos^{2 }A– sin^{2 }A = cos 2A so that the equation for the displacement of the particle is

*y* = 5 cos 6πt, which is in the form *y = A *cos *ωt*

Therefore, the angular frequency *ω* = 6π. The linear frequency *f* = *ω*/2π = 3 Hz.

Practice makes one perfect. So try to work out as many questions as possible to make you confident to face the AP Physics Exam. You will find some useful multiple choice questions in this section at physicsplus: Additional Multiple Choice Questions on Simple Harmonic Motion