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**MULTIPLE CHOICE QUESTIONS**** I**

**Q.1. The displacement of a particle is represented by the equationThe motion of the particle is(a) Simple harmonic with period 2p/w.(b) Simple harmonic with period π/ω.(c) Periodic but not simple harmonic.(d) Non-periodic.Ans.** (b)

- Acceleration is always directed towards the man position and so is always opposite to displacement, i.e. a ∝ -y or a = -ω
^{2}y - The differential equation of S.H.M is and the solution of this differential equation is y = a cos(ωt + φ
_{0})

**Method 1:** The displacement of the particle y = 3 cos

Velocity of the particle

Acceleration

Hence a = -4ω^{2}y = -(2ω)^{2}y

It means acceleration, a ∝ -y, the motion is SHM.

Hence angular frequency of S.H.M, ω' = 2ω

It means the motion is SHM with period π/ω.**Method 2:** Given the equation of displacement of the particle

We know cos(-θ) = cosθ

Hence y = 3cos ...(i)

Comparing with y = a cos(ωt + ϕ_{0})

Hence (i) represents simple harmonic motion with angular frequency 2ω.

Hence its time period, **Q.2. The displacement of a particle is represented by the equation y = sin ^{3}ωt. The motion is(a) Non-periodic.(b) Periodic but not simple harmonic.(c) Simple harmonic with period 2π/ω.(d) Simple harmonic with period π/ω.Ans.** (b)

There are certain motions that are repeated at equal intervals of time. Let the the interval of time in which motion is repeated. Then x(t) =x(t + T), where T is the minimum change in time. The function that repeats itself is known as a periodic function.

Given the equation of displacement of the particle, y = sin

We know sin 3θ = 3 sin θ - 4 sin3θ

Hence

⇒

⇒

⇒

⇒ d^{2}y/dt^{2} is not proportional to y.

Hence, motion is not SHM.

As the expression is involving sine function, hence it will be periodic.

Also sin^{3}ωt = (sin ωt)^{3}

= [sin(ωt + 2π)]^{3} = [sinω(t + 2π/ω)]^{3}

Hence, y = sin^{3}ωt represents a periodic motion with period 2π/ω.**Q.3. The relation between acceleration and displacement of four particles are given below:(a) a _{x} = + 2x.(b) a_{x} = + 2x^{2}.(c) a_{x }= – 2x^{2}.(d) a_{x} = – 2x.Which one of the particles is executing simple harmonic motion?Ans.** (d)

In case of simple harmonic motion, the acceleration is always directed towards the mean position and so is always opposite to displacement, i.e. a α = -x or a = -ω

In option (d) a

Hence it represents S.H.M.

(a) Periodic but not simple harmonic.

(b) Non-periodic.

(c) Simple harmonic and time period is independent of the density of the liquid.

(d) Simple harmonic and time-period is directly proportional to the density of the liquid.

Ans.

If the liquid in U-tube is filled to a height h and cross¬section of the tube is uniform and the liquid is incompressible and non- viscous. Initially the level of liquid in the two limbs will be at the same height equal to h. If the liquid is-pressed by y in one limb, it will rise by y along the length of the tube in the other limb, so the restoring force will be developed by hydrostatic pressure difference

Restoring force F = Weight of liquid column of height 2y

⇒ F = -(A *x* 2y *x* ρ) *x* g = -2Aρgy

⇒ F ∝ -y

Motion is SHM with force constant

k = 2Aρg

⇒ Timeperiod T =

which is independent of the density of the liquid.**Q.5. A particle is acted simultaneously by mutually perpendicular simple hormonic motions x = a cos ωt and y = a sin ωt . The trajectory of motion of the particle will be(a) An ellipse.(b) A parabola.(c) A circle.(d) A straight line.Ans.** (c)

If two S.H.M’s act in perpendicular directions, then their resultant motion is in the form of a straight line or a circle or a parabola etc. depending on the frequency ratio of the two S.H.Ms and initial phase difference. These figures are called Lissajous figures.

Let the equations of two mutually perpendicular S.H.M’s of same frequency be

x = a

Then the general equation of Lissajou's figure can be obtained as

This is a straight line which passes through origin and its slope is a_{2}/a_{1}.

Lissajou's Figures in other conditions

We have to find resultant - displacement by adding x and y-components. According to variation of x and y trajectory will be predicted.

Given, x = a cos ωt = a sin (ωt (π/2) ...(i)

And y = a sin ωt ...(ii)

Here a_{1} = a_{2} = a and ϕ = π/2

The trajectory of motion of the particle will be a circle of radius a.**Q.6. The displacement of a particle varies with time according to the relationy = a sin ωt + b cos ωt(a) The motion is oscillatory but not S.H.M.(b) The motion is S.H.M. with amplitude a + b.(c) The motion is S.H.M. with amplitude a** (d)

y = a sin ωt + b cos ωt

The amplitude of motion is given by

Hence, verifies the option (d).

**elastic support as A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,(a) D will vibrate with maximum amplitude.(b) C will vibrate with maximum amplitude.(c) B will vibrate with maximum amplitude.(d) All the four will oscillate with equal amplitude.Ans. **(b)

Here A is given a transverse displacement. Through the elastic support the disturbance is transferred to all the pendulums.

A and C are having same length, hence they will be in resonance, because of their time period of oscillation. Since length of pendulums A and C is same and T =2 π√L/g , hence their time period is same and they will have frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.

**(a) (b) (c) (d) Ans.** (a)

Suppose a particle P is moving uniformly on a circle of radius A with angular speed. Q and R are the two feets of the perpendicular drawn from P on two diameters one along .Y-axis and the other along Y-axis.

Suppose the particle P is on the X-axis at t = 0. Radius OP makes an angle with the X-axis at time t, then x = A cosωt and y = A sinωt

Here, x and v are the displacements of Q and R from the origin at time t, which are the displacement equations of SHM. It implies that although P is under uniform circular motion, Q and R are performing SHM about O with the same angular speed as that of P.

Let angular velocity of the particle executing circular motion is ω and when it is at P makes an angle θ as shown is the diagram.

As

Clearly, ω = ωt

x = B sin θ = B (sin ωt) = B sin(πt/15)

x = B sin (2π/30 (t))**Q.9. The equation of motion of a particle is x = a cos (α t) ^{2}. The motion is(a) Periodic but not oscillatory.(b) Periodic and oscillatory.(c) Oscillatory but not periodic.(d) Neither periodic nor oscillatory.Ans.** (c)

The equation of motion of a particle is

x = a cos(∝t)

is a cosine function and x varies between -a and +a, the motion is oscillatory. Now checking for periodic motion, putting t+T in place of t. T is supposed as period of the function ω(t).

x(t + T) = a cos [α(t + T)]

= a cos[α

Hence, it is not periodic.

(a) πs

(b) (π/2)s

(c) 2πs

(d) (π/t)s

Ans.

y = a sin ωt

v = dy/dt = aω cos ωt ...(i)

a = dv/dt = -aw sin ωt

a = -aω

v

v

v

∴ aω = 30 ...(iii)

a

60 = aω

60 = ω x 30

ω = 2 rad/s

2π/T = 2

T = π. Verifies option (a).

**(a) v _{1} + v_{2}.**(b)

(b)

(c)

(d)

Ans.

When a mass (m) is connected to a spring on a horizontal frictionless surface then their frequencies are

Now, the spring are in parallel because one end of both spring are connected with rigid body and other end to body. So their equivalent spring constant becomes k

And frequency

Form (i)

Hence, verifies the option (b).

**MULTIPLE CHOICE QUESTIONS II**

**Q.12. The rotation of earth about its axis is(a) Periodic motion.(b) Simple harmonic motion.(c) Periodic but not simple harmonic motion.(d) Non-periodic motion.Ans. **(a, c)

The motion of earth about its own axis is circular and complete its one complete revolution in regular interval of time. So it is periodic. But motion is not about a fixed point from which we can measure it’s displacement or about which it moves both side so it is not simple harmonic motion. So verifies the option (a) and (c).

(a) Simple harmonic motion.

(b) Non-periodic motion.

(c) Periodic motion.

(d) Periodic but not S.H.M.

Ans.

For small angular displacement, the situation is shown in the figure. Only one restoring force creates motion in ball inside bowl.

Only one restoring force creates motion in ball inside bowl.

F = -mg sinθ

As θ is small, sin θ = θ

So, ma = -mg(x/R)

or, a = -(g/R)x ⇒ a ∝ -x

So, motion of the ball is S.H.M and periodic.**Q.14. Displacement vs. time curve for a particle executing S.H.M is shown in figure. Choose the correct statements.**

**(a) Phase of the oscillator is same at t = 0 s and t = 2 s(b) Phase of the oscillator is same at t = 2 s and t = 6 s.(c) Phase of the oscillator is same at t = 1 s and t = 7 s.(d) Phase of the oscillator is same at t = 1 s and t = 5 s.Ans.** (b, d)

Two particles are said to be in same phase if the mode of vibration is same i.e., their distance will be nλ(/n = 1, 2, 3, ...)

(a) Distance between particles at t t = 0 and t = 2 is λ/2.

So, articles are not in same phase.

(b) As from figure the particles at t = 2 sec and 6 sec are at distance λ, so are in same phase.

(c) Particles at t =1, t= 7 are the distance so are not in phase.

(d) Particles at t = 1 sec and 5 sec are at distance = λ so are in same phase.**Q.15. Which of the following statements is/are true for a simple harmonic oscillator?(a) Force acting is directly proportional to displacement from the mean position and opposite to it.(b) Motion is periodic.(c) Acceleration of the oscillator is constant.(d) The velocity is periodic.Ans. **(a, b, d)

Consider a SHM x = a sin ωt. ...(i)

...(ii)

Acceleration

∴ A = -αω

mA = -mω

F ∝ -(x)

Hence, force is directly proportional to displacement opposite the direction of displacement. We can prove

x(t) = x(t+T)

So motion is periodic and SHM.

From (ii) v is also periodic as we can prove

v(t) = v(t + T)

Hence, verifies the option

**(a) The force is zero at t = 3T/4.(b) The acceleration is maximum at y = 4T/4.(c) The velocity is maximum at t = T/4.(d) The P.E. is equal to K.E. of oscillation at t = T/2.Ans. **(a, b, c)

(a) At t = 3T/4 Particle is at it’s mean position so force acting on it is zero, but it continue the motion due to inertia of mass, here a = 0, so F = 0.

(b) At t = 4T/4 = T, particle’s velocity changes increasing to decreases so maximum change in velocity at T. As acceleration = Change in velocity/Time, so acceleration is maximum here.

(c) At t = T/4 it is at its mean position where the velocity is maximum as no retarding force on it.

(d) t = T/2 = 2T/4, the particle has K.E = 0. So KE ≠ PE.

Hence, verifies the option (a,b,c)

(b) Average kinetic energy per cycle is equal to half of its maximum kinetic energy.

(c) Mean velocity over a complete cycle is equal to 2/π times of its maximum velocity.

(d) Root mean square velocity is 1/√2 times of its maximum velocity.

Ans.

In case of S.H.M, average total energy per cycle

= Maximum kinetic energy (K

= Maximum potential energy (U

Average K.E per cycle =

Let us write the equation for the SHM x = a sin ωt.

Clearly, it is a periodic motion as it involves sine function.

Let us find velocity of the particle,

Mean velocity over a complete cycle,

So,

Root mean square speed,

**(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive.(b) The sign of velocity of the particle at C going towards O is negative.(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative.(d) The sign of acceleration and force on the particle when it is at point B is negative.Ans.** (a, c, d)

**In option (a): **When the particle is 3 cm away from A going towards B. So, velocity is towards AB, i.e. positive. In SHM, acceleration is always towards mean position (O) it means both force and acceleration act towards O, have positive sign.

Hence option (a) is correct.**In option (b):** When the particle is at C, velocity is towards B hence positive. Hence option (b) is not correct.**In option (c):** When the particle is 4 cm away from B going towards A velocity is negative and acceleration is towards mean position (O), hence negative. Hence option (c) is correct.**In option (d):** Acceleration is always towards mean position (O). When the particle is at B, acceleration and force are towards BA that is negative. Hence option (d) is correct.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.1. Displacement versus time curve for a particle executing S.H.M. is shown in Figure, Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.**

**Ans.** In displacement-time graph of SHM, zero displacement values correspond to mean position; where velocity of the oscillator is maximum. Whereas the crest and troughs represent amplitude positions, where displacement is maximum and velocity of the oscillator is zero.

(i) The points A, C, E, G lie at extreme positions (maximum displacement, y = A). Hence the velocity of the oscillator is zero.

(ii) The points B, D, F, H lie at mean position (zero displacement, y = 0). We know the speed is maximum at mean position.**Q.2. Two identical springs of spring constant K are attached to a block of mass m and to fixed supports as shown in Fig. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force**

**Ans.** When mass is displaced from equilibrium position by a distance x towards right, the right spring gets compressed by x developing a restoring force kx towards left on the block. The left spring is stretched by an amount x developing a restoring force kx left on the block.

developing a restoring force kx towards left on the block.

F_{1} = -kx (for left spring) and

F_{2} = -kx (for right spring)

Restoring force, F = F_{1} + F_{2} = -2kx

∴ F = 2kx towards left.**Q.3. What are the two basic characteristics of a simple harmonic motion?Ans.** The two basic characteristics of a simple harmonic motion

(i) Acceleration is directly proportional to displacement.

(ii) The direction of acceleration is always towards the mean position, that is opposite to displacement.

Ans.

The restoring torque about the fixed point O is

τ = mgl sinθ

If θ is small angle in radians, then sin θ = 0

⇒ mglθ

In vector form τ ∝ θ

Hence, motion of a simple pendulum is SHM for small angle of oscillations.

**Q.5. What is the ratio of maxmimum acceleration to the maximum velocity of a simple harmonic oscillator?Ans. **Consider a SHM. x = A sin ωt

v = dx/dt = Aω cos ωt

For v

∴ v

a = dv/dt = -Aω

For a

a

∴ a

Ans.

Total distance travelled while it goes from O → P → O → Q → O

= OP + PO + OQ + QO = A + A + A + A = 4A

Amplitude = OP = A

Hence, ratio of distance and amplitude = 4A/A = 4**Q.7. In Figure, what will be the sign of the velocity of the point P′ , which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in anticlockwise direction.**

**Ans.** As the particle on reference circle moves in anti-clockwise direction. The projection will move from P’ to O towards left, i.e. from right to left, hence sign is negative.

**Q.8. Show that for a particle executing S.H.M, velocity and displacement have a phase difference of π/2.Ans.** Consider a SHM x = A sin ωt ...(i)

v = dx/dt = Aω cos ωt = Aω sin (90° + ωt)

(∵ sin (90° + θ) = cosθ)

∴ v = Aω sin (ωt + (π/2))

Phase of displacement from (i) is (ωt)

Phase of velocity from (ii) is (ωt + (π/2))

Hence, the phase difference = ωt +π/2 - ωt = π/2.

Ans.

...(i)

When, PE is plotted against displacement x, we will obtain a parabola.

When x = 0, PE = 0.

When x = PE = maximum

= 1/2 mω

The kinetic energy (KE) of a simple harmonic oscillator KE = 1/2 mv

But velocity of oscillator v =

⇒

or KE = 1/2mω

This is also parabola, if we plot KE against displacement x.

i.e., KE = 0 at x = ±A

and KE = 1/2 mω

Now, total energy of the simple harmonic oscillator = PE + KE [using Eqs. (i) and (ii)]

Which is a constant and independent of x.

Plotting under the above guidelines KE, PE and TE versus displacement x-graph as follows:

**Important point: **From the graph we note that potential energy or kinetic energy completes two vibrations in a time during which S.H.M. completes one vibration. Thus the frequency of potential energy or kinetic energy is double than that of S.H.M.**Q.10. The length of a second’s pendulum on the surface of Earth is 1m. What will be the length of a second’s pendulum on the moon?Ans.** A second's pendulum means a simple pendulum having time period T = 2s.

The time period of a simple pendulum T =

where, l = length of the pendulum and

g = acceleration due to gravity on surface of the earth.

Thus,

or

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