RESONANCE SOL090313 - 1

HINTS & SOLUTIONS

DATE : 09-03-2013COURSE NAME : VIKAAS (JA) & VIPUL (JB)

TEST - 11TARGET : JEE (IITs) 2014

COURSE CODE : CLINIC CLASSES PHYSICS - Simple Harmonic Motion

1. (B)

Sol. 100TT

= 1002

1

is not valid as is not small.

g2T1

g22T2

% change = 1

12T

TT 100

= 100)12( = 41.4.2. (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

When it passes equilibrium position for the first time t = 4T

Phase = 24T

.

T2

=

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

between 2nd and 3rd collision is 2T

where

T = 2 km

Time between 2nd and 3rd collision is

t = 2T

= km

4. (B)

Sol. KA

2a

3a2

=

2a

mg A = Kmg3

5. (A)

Sol. T = g/2 TT

= 21

= 2

1 .t

So, the fractional change in the time period of a pendulum onchanging 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 = 101

T = 2 k

= 210/1

)2(.2552 2

= 2 102.2.10 = 4 second Ans.

7. (B)

Sol. f0 = 21

mg

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

f0 = 21

)12/mL(2L

.g.m2

=

21

Lg6

when bottom half of the stick is cut off

f0 = 21

12)2/L(

2m

4L

.g.2m

2 = Lg12

21

= 2 f0 Ans.

8. (A)Sol. X1 = 4cost

X2 = 3sintso X = 4cost + 3sint

= 5

tsin53tcos

54

= 5 [cos37. cost + sin37. sint]= 5 cos(t 37)so phase diffrence between X1 and X2 is 37

9. (B)

Sol. Just after cutting the string extension in spring = kmg3

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

Amplitude of oscillation

A = kmg3

kmg

= kmg2

.

10. (C)

Sol. Spring constant K = 1.04.6

= 64 N/m.

Now T = 2 km

or 64m2

4

m = 1 kg

11. (B)Sol. Speed of block is maximum at mean position. At mean position

upper spring is extended and lower spring is compressed.

RESONANCE SOL090313 - 2

12. (B)Sol. Potential energy U = mV

U = (50x2 + 100) 102

F = dxdU

= (100x) 102

m2x = (100 102 ) x10 103 2x = 100 102 x 2 = 100, = 10

f = 2 = 2

10 =

5

13. (C)Sol. U = 2 20 x + 5x2

F = dxdU

= 20 10x

At equilibrium position ; F = 020 10x = 0 x = 2

Since particle is released at x = 3, therefore amplitude ofparticle is 5.

5

03

5

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

14. (D)

Sol.

We know that if the particle was at point B at t = 0; Thenequation of SHM will bex = A cos (t)As the phase difference between point O and P is 30, sothat between P and B is 60.And as the particle is moving towards left at t = 0, so it willbe leading the SHM x = Acos (t) by 60 Hence

x = A cos

3t

T2

15. (A)16. (A,B)17. (A,B,D)

Sol. = m

K = 10 rad/s

T = s1022

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

4 = 4102

= 20

s.

Time taken to cover 0.1 m is 204T s

Time taken to cover 21

0.1m is 32

4T

= s3032

4102

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

with amplitude = tan1

ga

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

extremeegl

=tan1ag

time period will be 2

22 ga

.

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

Displacement x = x1 + x2 = 4 sin 3

= 32 m.

Resulting Amplitude

A= 2881643/cos)4)(2(242 22 = 72 m

Maximum speed = A = 720 m/s

Maximum acceleration = A2 = 7200 m/s2

Energy of the motion = 21

m2 AA2 = 28 J Ans.

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

k = 1120 N/m5kg 2kg

By momentum conservation,(5 3) + (2 10) = (5 + 2)VV = 5 m/sec.

Now, by energy conservation

21

5 32 + 21

2 102 = 21 (5 + 2)V2 + 2

1kx2

Put V and k

xmax= m41

= 25 cm.

Also first maximum compression occurs at ;

t = 4T3

= 43

k2 = 4

311207102

= 56

3 sec.

(where reduced mass , = 21

21mm

mm

).

21. (A,B,C)22. (A,C,D)

Sol. Kmax = 21

KA2

Eavg = 21

KA2

Vrms = 2V0

, Vavg = 0V2