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. cost + sin37º. sint] = 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 = 4cost X2 = 3sint so X = 4cost + 3sint
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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