WBUT CS Solved Question Papers SHM

WBUT CS Solved Question Papers SHM



1.1.                                                                                                                                                                                                        On superimposing two mutually perpendicular simple harmonic motion, we get u circular Lissajous figures when their phase difference (<p) and amplitudes (P, Q) are as follows:                                            [WBUT 2005(DECEMBER)]

a) (p = 0, P -Q                          b )(p = ^-,P = Q                              c) (p — 0, P*Q

Answer: (b)

^f72. The amplitudes (a & b) of two superimposing mutually perpendicular simple harmonic motions having phase difference <p. To get a circular Lissajous figure.

K                                    71

a)cj) = 0;a = b b) <(> = — ; a = b c) <j> = — ; a # b d) <)> = 0 ; a * b

— 2

Answer: (b)                                                                                                                                                                                              [WBUT 2006(DECEMBER)]

Two mutually perpendicular oscillations with same frequency, amplitude but phase difference 8 will produce closed curve with non-zero area enclosed


a) for all values of 8 except 8 = 0                       b) only for 8 = —


c) for all values of 8 except 8 = 0 and 8 = 71 d) for all values of 8 > —


Answer: (c)                                                                                                                                              [WBUT 2007(DECEMBER), 2011 (JUNE)]

Two mutually perpendicular S.H.Ms with equal time periods but different ^amplitudes are superposed. If the phase difference between these oscillations it 45°, then they form a                     [WBUT 2008 (DECEMBER)]

a) circle            b) straight line         c) ellipse                          d) parabola

Answer: (c)

)£ 1.5. When a spring with spring constant K is cut into three equal parts, the force constant of each of the part would be [WBUT 2009 (DECEMBER), 2012(JUNE)] a) K/3           b) 3/C                    c) 1             d) AC

Answer: (b)                                                                                             ^

JK6. Superposition of two S.H.M.s of equal time period and equal amplitude with phase difference (p =/r/ 2 forms                    [WBUT 2009 (DECEMBER)]

a) circle            b) ellipse c) parabola

Answer: (a)

/f/7. If a particle is executing simple harmonic motion with frequency v then its potential energy                          [WBUT 2010 (DECEMBER)]

a) remains constant over time                 b) is oscillating with a frequency v

c) is oscillating with a frequency v/2 d) is oscillating with a frequency 2 v

Answer: (d)

lXI^8. Resultant of two perpendicular simple harmonic motions of equal frequency and equal amplitude but a phase difference of 71 radian is

a) a straight line                   b) an ellipse [WBUT 2011 (DECEMBER)]

c) a circle                          d) a spiral collapsing inward

Answer: (a)

To get a circular Lissajous figure, the phase difference {(p) and the amplitude (a and b) of two superimposing, mutually perpendicular SHMs are respectively



b) 0=—, a=b

d) aa*b 4

a) <p = 0, a = b

[WBUT 2012(JUNE)]







2.1. Calculate the time period of the liquid column of length / in a U-tube, if it is
depressed in one arm by x, d is the density of liquid and A is the
cross-sectional area of each arm of U-tube.                                                 [WBUT 2007(DECEMBER)]

Answer:                                                                                           *-

Let us assume a U- tube of uniform bore with the two ‘arms vertical is partly filled with a
liquid upto the level AB indicated by dotted line in. Mass of liquid in the tube is
m = LAp, where / = total length of liquid column p is the density of the liquid and A is

the area of cross-section of the tube. Now, the liquid in the left arm is depressed by a
small distance x. The liquid in the right arm will rise by the same height x as the area of
cross section of the tube is uniform and the liquid is incompressible. Hence, the
difference in height of liquid in the two arms of the tube is 2x.

Total unbalanced force acting on the whole of the liquid is

F/ = weight of the liquid column of height 2x.

F/ = 2xApg

When the liquid on the right arm is released force tends to
restore the liquid level in both the arms at the same position.

The liquid level is found to execute oscillatory motion.

So, the restoring force F = —2xApg


The acceleration of the liquid column f = —




which is the required time period and frequency of oscillation.

2.2.    A cubical block of side L cm and density d is floating in a water of density PiP > d). The block is slightly depressed and released. Show that it will execute

simple harmonic motion and hence determine the frequency of oscillation.

[WBUT 2009(DECEMBER), 2011(JUNE)]


As the cubical block of mass M is floating on the water of density p creates a static

equilibrium. The weight of the block is balanced by the weight of liquid (water) it displaces.

Let we depress the block by £ distance x (by dipping it further in the water), the buoyant force on the cubical block increases by pL2gx as L2 p x is the mass of the liquid

displaced by further dipping.

Neglecting viscous force we can write restoring force F = – pi}gx M^ = -pL2g x


. d2x _ pi}g

dt2 M

We observe that acceleration is directly proportional to displacement hence, the motion of the block is simple harmonic.

Comparing this equation with the general equation of S.H.M / = —CiTx we have


M                         V M

Now mass of the block is Ld

2.3 Derive the expression for thex = a sin COt total energy of a simple harmonic oscillator and show that it is constant and proportional to the square of the amplitude.                                                [WBUT 2011 (DECEMBER)]


Let x = a sin COt be the equation of a simple harmonic oscillator.

11                    1 ( I                                           ^

So kinetic energy K.E= —mv2 = — ma2C02 cos2 COt = —ma OT J1———————–

2 2                                       2               V a

K.E= — mv2 = — mar (a2 – x2); and potential energy P.E = — kx2 2 2 2

So, Total energy E = T + V = —mC02 (a2 -jc2) + — fct2 = —mO)2a2

2 2 2

Which is a constant quantity and proportional to amplitude ‘a’.

2.4. The displacement of any particle at any instant t is given by x-3 co-sfflf + 4 sin ox. Show that the motion is simple harmonic. What is the

amplitude of this oscillation? Show that its kinetic energy oscillates with angular

frequency 2CO.                                                                      [WBUT 2012(JUNE)]


Displacement of the particle is given by x = 3 cos 6* + 4 sin COt

Let us consider 3 = Asintp … (1) and 4 = Acostj) … (2). On substituting these values in the above expression we have x = A sin (p cos CtX + A cos (p sin COt, or, x = A sin ( COt + 0),

which is an equation of a simple harmonic motion having an amplitude A and phase

/ ^

(f) = tan 1

Now amplitude A can be found out by squaring and adding (1) and (2). So 9 + 16 = A2 —>A = 5unit.

From the expression of the displacement we have x = A sin (COt 4- <p)

So kinetic energy will be — — mA2ct? cos2 (ax + </>)

Further we can write,

K.E = -^-mAW2cos2 (ax + <p) = ^mA2af {cos 2(ax + <p) +1} or, K.E = ^mA2ajl                                    cos(2<af + 20)

Since mA2C02 is a constant term so, the first term of K.E is constant and the 2nd term varies with cos 2cot. Hence Proved.


2.5. The displacement of a simple harmonic oscillator is given by ;t=a sin {cot + 0). If the oscillations started at time t = 0 from a position x0 with velocity *=v0, show












































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