NEET Exam  >  NEET Notes  >  Physics Class 11  >  DC Pandey Solutions: Simple Harmonic Motion - 2

DC Pandey Solutions: Simple Harmonic Motion - 2 | Physics Class 11 - NEET PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


Introductory Exercise 11.2 
Q 1.  A simple pendulum of length l and mass m is suspended in a car that is moving with constant 
speed v around a circle of radius r. Find the period of oscillation and equilibrium position of the 
pendulum. 
Q 2.  Find the period of oscillation of a pendulum of length l if its point of suspension is. 
  (a) moving vertically up with acceleration a. 
  (b) moving vertically down with acceleration a (< g). 
  (c) falling freely under gravity 
  (d) moving horizontally with acceleration a. 
Q 3.  A clock with an iron pendulum keeps correct time at 20°C. How much time will it lose or gain in a 
day if the temperature changes to 40°C. Thermal coefficient of linear expansion ? = 0.0000] 2 per 
°C. 
Q 4.  A simple pendulum with a solid metal bob has a period T. What will be the period of the same 
pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the 
bob? 
Solutions 
1.  At equilibrium,  
     
   F
 
cos ? = mg  
   
   
2.  (a) g
eff
 = g + a 
    
  (b) g
eff
 = g – a 
    
  (c) g
eff
 = 0 ? T ? ?  
  (d)  
   
3.  
Page 2


Introductory Exercise 11.2 
Q 1.  A simple pendulum of length l and mass m is suspended in a car that is moving with constant 
speed v around a circle of radius r. Find the period of oscillation and equilibrium position of the 
pendulum. 
Q 2.  Find the period of oscillation of a pendulum of length l if its point of suspension is. 
  (a) moving vertically up with acceleration a. 
  (b) moving vertically down with acceleration a (< g). 
  (c) falling freely under gravity 
  (d) moving horizontally with acceleration a. 
Q 3.  A clock with an iron pendulum keeps correct time at 20°C. How much time will it lose or gain in a 
day if the temperature changes to 40°C. Thermal coefficient of linear expansion ? = 0.0000] 2 per 
°C. 
Q 4.  A simple pendulum with a solid metal bob has a period T. What will be the period of the same 
pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the 
bob? 
Solutions 
1.  At equilibrium,  
     
   F
 
cos ? = mg  
   
   
2.  (a) g
eff
 = g + a 
    
  (b) g
eff
 = g – a 
    
  (c) g
eff
 = 0 ? T ? ?  
  (d)  
   
3.  
   
  ?  With increase in temperature, pendulum clock becomes slow. 
    × 12 × 10
-6
 × 86400 × 20 
   = 1.2 × 8.64 s = 10.37 s 
4.  
     
    
    
 
Introductory Exercise 11.3 
Q 1.  Find the period of oscillation of the system shown in Fig.  
 
Q 2.  A block of mass 0.2 kg is attached to a massless spring of force constant 80 N/m as shown in Fig. 
Find the period of oscillation. Take g = 10 m/s
2
. Neglect friction. 
 
Q 3.  A bullet of mass m strikes a block of mass M. The bullet remains embedded in the block. Find the 
amplitude of the resulting SHM, 
 
Page 3


Introductory Exercise 11.2 
Q 1.  A simple pendulum of length l and mass m is suspended in a car that is moving with constant 
speed v around a circle of radius r. Find the period of oscillation and equilibrium position of the 
pendulum. 
Q 2.  Find the period of oscillation of a pendulum of length l if its point of suspension is. 
  (a) moving vertically up with acceleration a. 
  (b) moving vertically down with acceleration a (< g). 
  (c) falling freely under gravity 
  (d) moving horizontally with acceleration a. 
Q 3.  A clock with an iron pendulum keeps correct time at 20°C. How much time will it lose or gain in a 
day if the temperature changes to 40°C. Thermal coefficient of linear expansion ? = 0.0000] 2 per 
°C. 
Q 4.  A simple pendulum with a solid metal bob has a period T. What will be the period of the same 
pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the 
bob? 
Solutions 
1.  At equilibrium,  
     
   F
 
cos ? = mg  
   
   
2.  (a) g
eff
 = g + a 
    
  (b) g
eff
 = g – a 
    
  (c) g
eff
 = 0 ? T ? ?  
  (d)  
   
3.  
   
  ?  With increase in temperature, pendulum clock becomes slow. 
    × 12 × 10
-6
 × 86400 × 20 
   = 1.2 × 8.64 s = 10.37 s 
4.  
     
    
    
 
Introductory Exercise 11.3 
Q 1.  Find the period of oscillation of the system shown in Fig.  
 
Q 2.  A block of mass 0.2 kg is attached to a massless spring of force constant 80 N/m as shown in Fig. 
Find the period of oscillation. Take g = 10 m/s
2
. Neglect friction. 
 
Q 3.  A bullet of mass m strikes a block of mass M. The bullet remains embedded in the block. Find the 
amplitude of the resulting SHM, 
 
Q 4.  A spring is cat into three equal pieces and connected as shown in problem number (1) of the same 
exercise. By what factor will the time period of oscillation change if a block is attached before and 
after? 
Solutions 
1.  
   
2.  
    
3.  
   
    
4.  
    
    
  i.e., time period becomes times. 
 
Introductory Exercise 11.4 
Q 1.  An annular ring of internal and outer radii r and R respectively oscillates in a vertical plane about a 
horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate 
its time period and show that the length of an equivalent simple pendulum is 
3R
2
as r ? 0 and 2R 
as r ? R. 
Q 2.  A body of mass 200 g oscillates about a horizontal axis at a distance of 20 cm from its centre of 
gravity. If the length of the equivalent simple pendulum is 35 cm, find its moment of inertia about 
the point of suspension. 
Solutions 
Page 4


Introductory Exercise 11.2 
Q 1.  A simple pendulum of length l and mass m is suspended in a car that is moving with constant 
speed v around a circle of radius r. Find the period of oscillation and equilibrium position of the 
pendulum. 
Q 2.  Find the period of oscillation of a pendulum of length l if its point of suspension is. 
  (a) moving vertically up with acceleration a. 
  (b) moving vertically down with acceleration a (< g). 
  (c) falling freely under gravity 
  (d) moving horizontally with acceleration a. 
Q 3.  A clock with an iron pendulum keeps correct time at 20°C. How much time will it lose or gain in a 
day if the temperature changes to 40°C. Thermal coefficient of linear expansion ? = 0.0000] 2 per 
°C. 
Q 4.  A simple pendulum with a solid metal bob has a period T. What will be the period of the same 
pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the 
bob? 
Solutions 
1.  At equilibrium,  
     
   F
 
cos ? = mg  
   
   
2.  (a) g
eff
 = g + a 
    
  (b) g
eff
 = g – a 
    
  (c) g
eff
 = 0 ? T ? ?  
  (d)  
   
3.  
   
  ?  With increase in temperature, pendulum clock becomes slow. 
    × 12 × 10
-6
 × 86400 × 20 
   = 1.2 × 8.64 s = 10.37 s 
4.  
     
    
    
 
Introductory Exercise 11.3 
Q 1.  Find the period of oscillation of the system shown in Fig.  
 
Q 2.  A block of mass 0.2 kg is attached to a massless spring of force constant 80 N/m as shown in Fig. 
Find the period of oscillation. Take g = 10 m/s
2
. Neglect friction. 
 
Q 3.  A bullet of mass m strikes a block of mass M. The bullet remains embedded in the block. Find the 
amplitude of the resulting SHM, 
 
Q 4.  A spring is cat into three equal pieces and connected as shown in problem number (1) of the same 
exercise. By what factor will the time period of oscillation change if a block is attached before and 
after? 
Solutions 
1.  
   
2.  
    
3.  
   
    
4.  
    
    
  i.e., time period becomes times. 
 
Introductory Exercise 11.4 
Q 1.  An annular ring of internal and outer radii r and R respectively oscillates in a vertical plane about a 
horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate 
its time period and show that the length of an equivalent simple pendulum is 
3R
2
as r ? 0 and 2R 
as r ? R. 
Q 2.  A body of mass 200 g oscillates about a horizontal axis at a distance of 20 cm from its centre of 
gravity. If the length of the equivalent simple pendulum is 35 cm, find its moment of inertia about 
the point of suspension. 
Solutions 
1.  
    
    
   = ? ? (R
2 
- r
2
) 
    
    
    
   
  and for r ? R ? l
eff
 = 2R 
2.  Here, l
eff
 = 35 cm and l =20 cm 
    
  ?  200 g × 20 cm × 35 cm = 1.4 × 10
5
 g-cm
2
 
 
Introductory Exercise 11.5 
Q 1.  A particle is subjected to two simple harmonic motions of the same frequency and direction. The 
amplitude of the first motion is 4.0 cm and that of the second is 3.0 cm. Find the resultant 
amplitude if the phase difference between the two motions is: 
  (a) 0°    (b) 60°   (c) 90°   (d) 180° 
Q 2.  A particle is subjected to two simple harmonic motions. 
Page 5


Introductory Exercise 11.2 
Q 1.  A simple pendulum of length l and mass m is suspended in a car that is moving with constant 
speed v around a circle of radius r. Find the period of oscillation and equilibrium position of the 
pendulum. 
Q 2.  Find the period of oscillation of a pendulum of length l if its point of suspension is. 
  (a) moving vertically up with acceleration a. 
  (b) moving vertically down with acceleration a (< g). 
  (c) falling freely under gravity 
  (d) moving horizontally with acceleration a. 
Q 3.  A clock with an iron pendulum keeps correct time at 20°C. How much time will it lose or gain in a 
day if the temperature changes to 40°C. Thermal coefficient of linear expansion ? = 0.0000] 2 per 
°C. 
Q 4.  A simple pendulum with a solid metal bob has a period T. What will be the period of the same 
pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the 
bob? 
Solutions 
1.  At equilibrium,  
     
   F
 
cos ? = mg  
   
   
2.  (a) g
eff
 = g + a 
    
  (b) g
eff
 = g – a 
    
  (c) g
eff
 = 0 ? T ? ?  
  (d)  
   
3.  
   
  ?  With increase in temperature, pendulum clock becomes slow. 
    × 12 × 10
-6
 × 86400 × 20 
   = 1.2 × 8.64 s = 10.37 s 
4.  
     
    
    
 
Introductory Exercise 11.3 
Q 1.  Find the period of oscillation of the system shown in Fig.  
 
Q 2.  A block of mass 0.2 kg is attached to a massless spring of force constant 80 N/m as shown in Fig. 
Find the period of oscillation. Take g = 10 m/s
2
. Neglect friction. 
 
Q 3.  A bullet of mass m strikes a block of mass M. The bullet remains embedded in the block. Find the 
amplitude of the resulting SHM, 
 
Q 4.  A spring is cat into three equal pieces and connected as shown in problem number (1) of the same 
exercise. By what factor will the time period of oscillation change if a block is attached before and 
after? 
Solutions 
1.  
   
2.  
    
3.  
   
    
4.  
    
    
  i.e., time period becomes times. 
 
Introductory Exercise 11.4 
Q 1.  An annular ring of internal and outer radii r and R respectively oscillates in a vertical plane about a 
horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate 
its time period and show that the length of an equivalent simple pendulum is 
3R
2
as r ? 0 and 2R 
as r ? R. 
Q 2.  A body of mass 200 g oscillates about a horizontal axis at a distance of 20 cm from its centre of 
gravity. If the length of the equivalent simple pendulum is 35 cm, find its moment of inertia about 
the point of suspension. 
Solutions 
1.  
    
    
   = ? ? (R
2 
- r
2
) 
    
    
    
   
  and for r ? R ? l
eff
 = 2R 
2.  Here, l
eff
 = 35 cm and l =20 cm 
    
  ?  200 g × 20 cm × 35 cm = 1.4 × 10
5
 g-cm
2
 
 
Introductory Exercise 11.5 
Q 1.  A particle is subjected to two simple harmonic motions of the same frequency and direction. The 
amplitude of the first motion is 4.0 cm and that of the second is 3.0 cm. Find the resultant 
amplitude if the phase difference between the two motions is: 
  (a) 0°    (b) 60°   (c) 90°   (d) 180° 
Q 2.  A particle is subjected to two simple harmonic motions. 
   x
1
 = 4.0 sin(100 ?t) and x
2
 = 3.0 sin 100 t
3
? ??
??
??
??
 
  Find ; 
  (a) the displacement at t = 0 
  (b) the maximum speed of the particle and 
  (c) the maximum acceleration of the particle. 
Solutions 
1.  Here, a
1
 = 4.0 cm and  
   a
2
 =3.0 cm  
  (a) For, ? ?
 
= 0, a = a
max
 = a
1
 + a
2
 
   = 4.0 + 3.0 = 7.0 cm  
  (b) For, ? ?
 
= 60°, 
    
  (c) For, ? ? = 90°  
    
  (d) For, ? ?
 
= 180°, a = a
min 
= a
1 
- a
2  
   
= 4.0 - 3.0 = 10 cm 
2.  x = x
1
 + x
2
 = 4 sin (100 ?t) 
      
    
    
    
  (a) At t = 0, x = 4 sin 0 + 3 sin 
3
?
 
    
  (b)  
  (c) 
Read More
98 videos|388 docs|105 tests

Top Courses for NEET

FAQs on DC Pandey Solutions: Simple Harmonic Motion - 2 - Physics Class 11 - NEET

1. What is simple harmonic motion?
Ans. Simple harmonic motion (SHM) is a type of periodic motion where an object oscillates back and forth around its equilibrium position due to a restoring force that is directly proportional to its displacement. It can be described by a sinusoidal function and is commonly observed in systems like pendulums, springs, and vibrating strings.
2. How is the period of simple harmonic motion calculated?
Ans. The period of simple harmonic motion can be calculated using the formula T = 2π√(m/k), where T represents the period, π is a mathematical constant (approximately 3.14), m is the mass of the object undergoing SHM, and k is the force constant or the spring constant.
3. What is the amplitude in simple harmonic motion?
Ans. The amplitude in simple harmonic motion refers to the maximum displacement of an object from its equilibrium position during one complete oscillation. It represents the maximum value of the displacement and determines the energy of the oscillating system. A larger amplitude corresponds to a greater energy transfer.
4. How does the frequency relate to the period in simple harmonic motion?
Ans. The frequency of simple harmonic motion is the number of complete oscillations or cycles that occur in one second. It is inversely related to the period of the motion, which is the time taken for one complete oscillation. The frequency can be calculated using the formula f = 1/T, where f represents the frequency and T is the period.
5. What is the role of the restoring force in simple harmonic motion?
Ans. The restoring force in simple harmonic motion is responsible for bringing the object back to its equilibrium position when it is displaced. It is directly proportional to the displacement and acts in the opposite direction to the displacement, resulting in oscillatory motion. The restoring force is usually provided by a spring, gravity, or any other force that follows Hooke's law.
98 videos|388 docs|105 tests
Download as PDF
Explore Courses for NEET exam

Top Courses for NEET

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Viva Questions

,

ppt

,

practice quizzes

,

Sample Paper

,

past year papers

,

Semester Notes

,

DC Pandey Solutions: Simple Harmonic Motion - 2 | Physics Class 11 - NEET

,

Exam

,

Previous Year Questions with Solutions

,

Objective type Questions

,

study material

,

pdf

,

Important questions

,

mock tests for examination

,

shortcuts and tricks

,

Extra Questions

,

DC Pandey Solutions: Simple Harmonic Motion - 2 | Physics Class 11 - NEET

,

video lectures

,

DC Pandey Solutions: Simple Harmonic Motion - 2 | Physics Class 11 - NEET

,

MCQs

,

Summary

,

Free

;