Physics - Rigid Body Dynamics solutions

TEST - 11 DATE : 09-03-2013

COURSE NAME : VIKAAS (JA) & VIPUL (JB)

TARGET : JEE (IITs) 2014 COURSE CODE : CLINIC CLASSES PHYSICS - Simple Harmonic Motion

HINTS & SOLUTIONS 1. Sol.

7.

(B)

1  T  100 =  100 is not valid as  is not small. T 2 

T1  2

 g

T2  2

2 g

Sol. f 0 =

= ( 2 – 1)  100 = 41.4.

1 f0 = 2

2. (B) Sol. Phase of the motion is (t + ). Using x = A sin (t + ) and V = A cos(t + ) for conditions at t = 0  x = A and V = 0 then  = /2

Phase =

t=

4.

m L .g. 1 12g 2 4  = = m (L / 2)2 2 L 2 12

2 f0

An s .

3 4   5 cos t  5 sin t   

= 5 [cos37º. cost + sin37º. sint] = 5 cos(t – 37º) so phase diffrence between X1 and X2 is 37º 9.

m T = 2 k

(B)

Sol. Just after cutting the string extension in spring =

(B)

3mg k

The extension in the spring when block is in mean position =

Sol. KA × 5.

=5

m k

t=

6g L

8. (A) Sol. X1 = 4cost X2 = 3sint so X = 4cost + 3sint

T where 2

Time between 2nd and 3rd collision is



L 2 = 1 2 2 (mL / 12) m.g.

1 f 0’ = 2

T 4

3. (B) Sol. Time period is independent of amplitute in SHM. Hence the time

T = 2

mg  

when bottom half of the stick is cut off

2 T  .  = T 4 2

between 2nd and 3rd collision is

1 2

where,  is distance between point of suspension and centre of mass of the body. Thus, for the stick of length L and mass m :

T2  T1 % change = T1 × 100

When it passes equilibrium position for the first time

(B)

 2a a  a   = mg     3 2 2



A=

3mg K

mg k

(A)

Sol. T =

2  / g



1  1 T = =  .t T 2  2

So, the fractional change in the time period of a pendulum on changing the temperature is independent of length of pendulum. 6. (D) Sol. (D)  = – k 0.1 = – k(1.0), where k is torsional constant of the wire. k=

1 10



Amplitude of oscillation A=

3mg mg 2mg – = . k k k

10. (C) Sol. Spring constant K =

 T = 2 = 2 k = 2

2  25  (.2)2 5 1/ 10

10  .2  .2  10

= 4 second An s .

RESONANCE

Now T = 2

6.4 = 64 N/m. 0 .1

m  m  2 or  m = 1 kg 4 64 k

11. (B) Sol. Speed of block is maximum at mean position. At mean position upper spring is extended and lower spring is compressed.

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12. (B) Sol. Potential energy U = mV  U = (50x2 + 100) 10–2 F=



19. (A,B,C,D) Sol. At t = 0

dU = – (100x) 10–2 dx

Displacement x = x1 + x2

10 × 10–3  2x = 100 × 10–2 x  f=



 2 = 100,  = 10

2 2  4 2  2(2)( 4) cos  / 3  4  16  8  28

A=

 10 5 = = 2 2 

=

2 7 m

Maximum speed = A =

13. (C) Sol. U = 2 – 20 x + 5x2

20 7 m/s

Maximum acceleration = A 2 =

dU F=– = 20 – 10x dx At equilibrium position ; F = 0 20 – 10x = 0  x=2 Since particle is released at x = – 3, therefore amplitude of particle is 5.

5

 = m. 3 2 3

Resulting Amplitude

m 2x = – (100 × 10–2 ) x



= 4 sin

Energy of the motion =

200 7 m/s 2

1 m 2 A 2 = 28 J Ans. 2

20. (B,C,D) Sol. At max. extension both should move with equal velocity.

5 k = 1120 N/m

0

–3

2

5kg

7

It will oscillate about x = 2 with an amplitude of 5.  maximum value of x will be 7.

2kg

By momentum conservation, (5 × 3) + (2 × 10) = (5 + 2)V V = 5 m/sec. Now, by energy conservation 

14. (D) Sol. We know that if the particle was at point B at t = 0; Then equation of SHM will be x = A cos (t) As the phase difference between point O and P is 30º, so that between P and B is 60º. And as the particle is moving towards left at t = 0, so it will be leading the SHM x = Acos (t) by 60º Hence

  2 t  T 3  

x = A cos 

1 1 1 1 2 5 × 32 + × 2 × 102 = (5 + 2)V 2 + kx 2 2 2 2 Put V and k 

1 m = 25 cm. 4

Also first maximum compression occurs at ; t=

15. (A)

xmax=

 3T 3 2 = 4 4 k

=

10 3 3 2 = sec. 56 7  1120 4

16. (A,B) (where   reduced mass ,  =

17. (A,B,D) Sol.  =

K m T=

m1m 2 m1  m 2 ).

21. (A,B,C)

= 10 rad/s

22. (A,C,D)

2 2  s  10

Sol. Kmax =

1 KA2 2

Eavg =

1 KA2 2

Maximum speed will be at the natural length of the spring T/

 2 4= = s. 10  4 20 Time taken to cover 0.1 m is

Time taken to cover

T   s 4 20

Vrms =

V0 2

, Vavg =

2V0 

1  2 2 T 2   s × 0.1m is  = 10  4 3 30 4 3 2

18. (B,C) Sol. bob will oscillate about equilibrium position

with amplitude  = tan–1

a   g

a

–1

=tan g

for any value of a. If a < < g, motion will be SHM, and then

egl extreme

 time period will be 2 

RESONANCE

a 2  g2

.

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