Oscillations Practice Questions - DPP for JEE

 Page 1


1. (c)
Force applied by the spring is F = – kx
?  F = – k (2L?)
(? is the angular displacement from the equilibrium position). Further
Also, 
? 
? 
2. (c) When the bob moves from maximum angular displacement ? to
mean position, then the loss of gravitational potential energy is
mgh
where h = l(1 – cos ?)
3. (a) K. E. = 
At mean position, y = 0
 K. E. = 
P. E. = 
At , y = ; P. E. =  = 
 Ratio = 
Page 2


1. (c)
Force applied by the spring is F = – kx
?  F = – k (2L?)
(? is the angular displacement from the equilibrium position). Further
Also, 
? 
? 
2. (c) When the bob moves from maximum angular displacement ? to
mean position, then the loss of gravitational potential energy is
mgh
where h = l(1 – cos ?)
3. (a) K. E. = 
At mean position, y = 0
 K. E. = 
P. E. = 
At , y = ; P. E. =  = 
 Ratio = 
= 
4. (a) The two springs are in parallel.
 Effective spring constant,  K = K
1
 + K
2
Now, frequency of oscillation is given by
 or, f  = ....(i)
When both k
1
 and k
2
 are made four times their original values, the new
frequency is given by
= 2 f ; from eqn. (i)
5. (a) ; 
Net force 
t = 2t
0
Page 3


1. (c)
Force applied by the spring is F = – kx
?  F = – k (2L?)
(? is the angular displacement from the equilibrium position). Further
Also, 
? 
? 
2. (c) When the bob moves from maximum angular displacement ? to
mean position, then the loss of gravitational potential energy is
mgh
where h = l(1 – cos ?)
3. (a) K. E. = 
At mean position, y = 0
 K. E. = 
P. E. = 
At , y = ; P. E. =  = 
 Ratio = 
= 
4. (a) The two springs are in parallel.
 Effective spring constant,  K = K
1
 + K
2
Now, frequency of oscillation is given by
 or, f  = ....(i)
When both k
1
 and k
2
 are made four times their original values, the new
frequency is given by
= 2 f ; from eqn. (i)
5. (a) ; 
Net force 
t = 2t
0
6. (d) K.E = 
and P.E. = 
At mean position  d = 0. At extrement positions d = A
7. (b) For block A to move in S.H.M.
mg – N = m?
2
x
where x is the distance from mean position
For block to leave contact N = 0
8. (c) As we know, time period, T 
When additional mass M is added then
  or 
or, 
9. (d) Let the extension in the spring be x
0
 at equilibrium. If F
0
 be the
tension in the string then F
0
 = kx
0
. Further if T
0
 is the tension in the
Page 4


1. (c)
Force applied by the spring is F = – kx
?  F = – k (2L?)
(? is the angular displacement from the equilibrium position). Further
Also, 
? 
? 
2. (c) When the bob moves from maximum angular displacement ? to
mean position, then the loss of gravitational potential energy is
mgh
where h = l(1 – cos ?)
3. (a) K. E. = 
At mean position, y = 0
 K. E. = 
P. E. = 
At , y = ; P. E. =  = 
 Ratio = 
= 
4. (a) The two springs are in parallel.
 Effective spring constant,  K = K
1
 + K
2
Now, frequency of oscillation is given by
 or, f  = ....(i)
When both k
1
 and k
2
 are made four times their original values, the new
frequency is given by
= 2 f ; from eqn. (i)
5. (a) ; 
Net force 
t = 2t
0
6. (d) K.E = 
and P.E. = 
At mean position  d = 0. At extrement positions d = A
7. (b) For block A to move in S.H.M.
mg – N = m?
2
x
where x is the distance from mean position
For block to leave contact N = 0
8. (c) As we know, time period, T 
When additional mass M is added then
  or 
or, 
9. (d) Let the extension in the spring be x
0
 at equilibrium. If F
0
 be the
tension in the string then F
0
 = kx
0
. Further if T
0
 is the tension in the
thread then T
0
 = mg and  2T
0
 = kx
0
.
Let the mass m be displaced through a slight displacement x downwards.
Let the the new tension in the string and spring be T and F
respectively.
?    and F = 2T
?  
?   
?   Time period 
10. (a) The displacement of a particle in S.H.M. is given by
y = a sin (?t + f)
velocity =  = ?a cos (?t + f)
The velocity is maximum when the particle passes through the mean
position i.e.,
 = ? a
The kinetic energy at this instant is given by
m  =  m?
2
 a
2
 = 8 × 10
–3
 joule
or × (0.1) ?
2
 × (0.1)
2
 = 8 × 10
–3
Solving we get ? = ± 4
Substituting the values of a, ? and f in the equation of S.H.M., we get
y = 0.1 sin (± 4t + p/4) metre.
Page 5


1. (c)
Force applied by the spring is F = – kx
?  F = – k (2L?)
(? is the angular displacement from the equilibrium position). Further
Also, 
? 
? 
2. (c) When the bob moves from maximum angular displacement ? to
mean position, then the loss of gravitational potential energy is
mgh
where h = l(1 – cos ?)
3. (a) K. E. = 
At mean position, y = 0
 K. E. = 
P. E. = 
At , y = ; P. E. =  = 
 Ratio = 
= 
4. (a) The two springs are in parallel.
 Effective spring constant,  K = K
1
 + K
2
Now, frequency of oscillation is given by
 or, f  = ....(i)
When both k
1
 and k
2
 are made four times their original values, the new
frequency is given by
= 2 f ; from eqn. (i)
5. (a) ; 
Net force 
t = 2t
0
6. (d) K.E = 
and P.E. = 
At mean position  d = 0. At extrement positions d = A
7. (b) For block A to move in S.H.M.
mg – N = m?
2
x
where x is the distance from mean position
For block to leave contact N = 0
8. (c) As we know, time period, T 
When additional mass M is added then
  or 
or, 
9. (d) Let the extension in the spring be x
0
 at equilibrium. If F
0
 be the
tension in the string then F
0
 = kx
0
. Further if T
0
 is the tension in the
thread then T
0
 = mg and  2T
0
 = kx
0
.
Let the mass m be displaced through a slight displacement x downwards.
Let the the new tension in the string and spring be T and F
respectively.
?    and F = 2T
?  
?   
?   Time period 
10. (a) The displacement of a particle in S.H.M. is given by
y = a sin (?t + f)
velocity =  = ?a cos (?t + f)
The velocity is maximum when the particle passes through the mean
position i.e.,
 = ? a
The kinetic energy at this instant is given by
m  =  m?
2
 a
2
 = 8 × 10
–3
 joule
or × (0.1) ?
2
 × (0.1)
2
 = 8 × 10
–3
Solving we get ? = ± 4
Substituting the values of a, ? and f in the equation of S.H.M., we get
y = 0.1 sin (± 4t + p/4) metre.
11. (b)
Let BP = a, ?  x = OM = a sin ? = a sin (2t)
Hence M executes SHM within the given time period and its acceleration
is opposite to x that means towards left.
12. (a) Applying linear momentum conservation,
0.5 × 3 = (1 + 0.5) v   or v = 1 m/s
By conservation of energy,
After collision