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Wednesday, August 18, 2021

Oscillations And Waves : Physics

 


Q.1.  A particle is executing simple harmonic motion . Its displacement is given by x = 5 sin pi t. Where x is in cm and t in seconds. How long will the particle taken to move from the position of equilibrium to the position of maximum displacement ?

 A) 0.5 s

 B) 1.0 s

 C) 1.5 s

 D) 2.0 s

Explanation- 

Maximum displacement = amplitude = 5 cm At time t = 0 , x = 0 ( equilibrium position ) . Hence time t taken by the particle to move from x = 0 to x = 5 cm is given by space space space space space 5 equals 5 space sin space pi space t
o r space 1 equals space sin space pi space t space
o r space pi space t space equals space pi over 2 space o r space t space equals space 0.5 space s  


Q.2. The displacement at an instant 't', of particle executing a linear S.H.M. is given by

 x = 5 sin 31.4(t + 0.1). Its periodic time is 

A) 2 sec 

B) 0.2 sec 

C) 0.5 sec 

D) 1 sec 

Explanation-

straight x equals space 5 space sin space 31.4 space left parenthesis straight t plus 0.1 right parenthesis space equals 5 space sin space left square bracket 31.4 straight t plus 3.14 right square bracket space comparing space with space
straight x space equals space straight A space sin space open parentheses fraction numerator 2 πt over denominator straight T end fraction plus straight alpha close parentheses space we space get space fraction numerator 2 straight pi over denominator straight T space end fraction equals space 31.4
therefore space straight T space equals space fraction numerator 2 cross times 3.14 over denominator 31.4 end fraction equals space 0.2 space straight s space space 


Q.3. A particle is in UCM. From its every position a perpendicular is dropped. The point of intersection of the perpendicular and a diameter is called foot of the perpendicular and its motion is SHM...

A) Only on horizontal diameter

B) Only on vertical diameter

C) On only horizontal and vertical diameters

D) On any diameter

Explanation-

The motion on any diameter is SHM.


Q.4. The potential energy of a particle executing simple harmonic motion at a distance x from the equilibrium position is proportional to  

A) square root of x

B)  

C) x squared

D) x cubed 

Explanation-

The potential energy of a particle of mass m executing simple harmonic motion of angular frequency omega at a distance x from the equilibrium position is given by straight P. straight E. space equals space 1 half space mω squared straight x squared equals 1 half kx squared space where space straight k space equals space mω squared  is constant.


Q.5. A wave represented by the equation y=a cos( kx -omega t )b is superposed with another wave to form a stationary wave such that the point x=0 is a node. The equation of the other waves is

A) y apostrophe equals a space sin left parenthesis k x space plus omega t right parenthesis

B) y apostrophe equals negative a space cos space left parenthesis k x space minus space omega t right parenthesis

Cy apostrophe equals space minus space a cos left parenthesis k x plus space omega t right parenthesis

D) y apostrophe equals space minus a space sin space left parenthesis k x space minus space omega t right parenthesis 

Explanation-

To from a stationary wave, waves y and y' must travel in opposite directions. Wave y = a cos ( kx = straight omegat ) travels along the position x - direction. Waves y' = - a cos ( kx -  straight omegat ) and y' = - a sin ( kx - straight omegat ) in choices ( b ) and ( d ) are not possible. Choice ( a ) is also incorrect because at x = 0

y' = a sin straight omegat and y = a cos ( - straight omegat ) = a cos straight omegaTherefore, the resultant displacement at x = 0 which  is y + y' = a sin straight omegat +a cos straight omegat is not zero, i.e. these waves do not produce a node at x = 0. 


Q.6. Three sound waves of equal amplitudes have frequencies (v-1), (v) and (v+1). They superpose to give beats. The number of beats produced per second will be

A) v

B) v over 2

C) 2

D) 1

Explanation-

When the three waves superpose at a point, then from the superposition principle, the resultant particle displacement at that point is given by 

straight y equals straight y subscript 1 plus straight y subscript 2 plus straight y subscript 3
equals straight a space sin space open curly brackets 2 straight pi open parentheses straight v minus 1 close parentheses straight t close curly brackets plus straight a space sin space open parentheses 2 straight pi space straight v space straight t close parentheses space plus space straight a space sin space open curly brackets 2 straight pi open parentheses straight v plus 1 close parentheses straight t close curly brackets
Now space sin space open curly brackets 2 straight pi open parentheses straight v minus 1 close parentheses close curly brackets plus sin space open curly brackets 2 straight pi open parentheses straight v plus 1 close parentheses straight t close curly brackets space equals space 2 space cos space 2 space straight pi space straight t space sin space 2 space πvt
Therefore space comma space straight gamma equals straight a space open parentheses 1 plus 2 space cos space 2 space straight pi space straight t close parentheses space sin space 2 space πvt
or space straight y space equals space straight A space sin space 2 space straight pi space straight v space straight t space
where space straight A space equals space straight a space open parentheses 1 plus 2 space cos space 2 space straight pi space straight t close parentheses

is the resultant amplitude . Now , the resultant intensity straight A squared proportional to. space Now space straight A squared  will be maximum when cos 2 pi t = +1 

or 2 pi t = 0, 2pi , 4pi, …. etc.

or t = 0 , 1 s , 2 s , … etc. . 

therefore Time period of beats = time interval between two consecutive maxima = 1 s . Hence the beat frequency is 1 Hz.


Q.7. The differential equation of SHM of a seconds pendulum oscillating along  x axis is

A) fraction numerator straight d squared straight x over denominator dt squared end fraction = negative πx

B) fraction numerator straight d squared straight x over denominator dt squared end fraction space equals space πx

Cfraction numerator straight d squared straight x over denominator dt squared end fraction space equals space minus straight pi squared straight x

D) fraction numerator straight d squared straight x over denominator dt squared end fraction space equals space straight pi squared straight x 

Explanation-

For seconds pendulum

T = 2 sec              therefore space straight omega space equals space fraction numerator 2 straight pi over denominator straight T end fraction space equals space fraction numerator 2 straight pi over denominator 2 end fraction space equals space straight pi

The diff. eqn. off SHM is

fraction numerator straight d squared straight x over denominator dt squared end fraction space equals space minus straight omega squared straight x space or space space fraction numerator straight d squared straight x over denominator dt squared end fraction space equals space minus straight pi squared straight x 


Q.8. Two SHMs of same period and amplitudes straight a subscript 1 and straight a subscript 2 act parallel to each other on a particle and have resultant amplitude straight A subscript 1 when the phase difference between them is 180 to the power of degree the resultant amplitude is straight A subscript 2. The amplitudes  straight a subscript 1and straight a subscript 2 respectively are...

A) left parenthesis straight A subscript 1 space plus space straight A subscript 2 right parenthesis comma space left parenthesis straight A subscript 1 space minus space straight A subscript 2 right parenthesis

B) left parenthesis straight A subscript 1 space plus space straight A subscript 2 right parenthesis divided by straight A subscript 1 straight A subscript 2 comma end subscript space left parenthesis straight A subscript 1 space minus space straight A subscript 2 right parenthesis divided by straight A subscript 1 straight A subscript 2

C) straight A subscript 1 space straight A subscript 2 divided by space left parenthesis straight A subscript 1 space plus space straight A subscript 2 right parenthesis comma space straight A subscript 1 straight A subscript 2 divided by left parenthesis straight A subscript 1 space minus space straight A subscript 2 right parenthesis

Dleft parenthesis straight A subscript 1 space plus space straight A subscript 2 right parenthesis divided by 2 comma space left parenthesis straight A subscript 1 space minus space straight A subscript 2 right parenthesis divided by 2 

Explanation-

When phase diff. empty set space equals space 0

R subscript m a x space equals space end subscript a space plus space a subscript 2
S o space A subscript 1 space equals space a subscript 1 space plus space a subscript 2
w h e n space empty set space equals space 180 to the power of degree space equals space straight pi comma space space space space space space straight R subscript min space equals space straight a subscript 1 space minus space straight a subscript 2 space equals space straight A subscript 2
Adding space straight A subscript 1 space plus space straight A subscript 2 space equals space 2 straight a subscript 1 space space space or space straight a subscript 1 space equals space fraction numerator straight A subscript 1 space plus space straight A subscript 2 over denominator 2 end fraction
Substracting space straight a subscript 2 space equals space fraction numerator straight A subscript 1 space minus space straight A subscript 2 over denominator 2 end fraction 


Q.9.  A travelling wave in a stretched string is described by the equation y equals A space sin space left parenthesis thin space k x space minus space omega t space right parenthesis The maximum particle velocity is 

AA omega

B) omega over k

C) fraction numerator d space omega over denominator d space k end fraction

D) x over t 

Explanation-

Particle velocity V = dy over dt equals fraction numerator straight d over denominator d t end fraction open square brackets straight A space sin space left parenthesis space kx space minus space ωt space right parenthesis space close square brackets space equals space A omega space cos space left parenthesis space k x minus omega t space right parenthesis
Hecne space straight V subscript max space end subscript equals Aω space comma space which space is space choice space left parenthesis thin space straight a space right parenthesis. 


Q.10.  A simple pendulum attached to the ceiling of a stationary lift has a time period T. Te distance y covered the lift moving upwards varies with time t as y = t squared where y is in meter and t in second. If g = 10 msblank to the power of negative 2 end exponent, the time period of the pendulum will be 

A) square root of 4 over 5 end root T

Bsquare root of 5 over 6 end root T

C) square root of 5 over 4 end root T

D) square root of 6 over 5 end root T

Explanation-

Given y = tblank squared. The velocity of the lift varies with t as 

ϑ equals dy over dt equals 2 straight t

therefore Acceleration a = dϑ over dt equals 2 space m s to the power of negative 2 end exponent , directed upwards , Hence 

straight T apostrophe equals 2 space straight pi space square root of fraction numerator straight l over denominator straight g plus straight a end fraction end root
and space straight T equals 2 space straight pi space square root of straight l over straight g end root
therefore space fraction numerator straight T apostrophe over denominator straight T end fraction equals square root of fraction numerator straight g over denominator straight g plus straight a end fraction end root space equals space square root of fraction numerator 10 over denominator open parentheses 10 plus 2 close parentheses end fraction end root space equals space square root of 5 over 6 end root


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