Download, print and study this document offline |
Page 1 Previous Year Questions (2021-24): Simple Harmonic Motion (Oscillations) 2024 Q1 - 2024 (01 Feb Shift 2) A mass ?? is suspended from a spring of negligible mass and the system oscillates with a frequency ?? 1 . The frequency of oscillations if a mass 9 m is suspended from the same spring is f 2 . The value of f 1 f 2 is Q2 - 2024 (27 Jan Shift 1) A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is v ?? cm, where ?? = Q3 - 2024 (27 Jan Shift 2) A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ( ?? ) of thread deflection in the extreme position will be : (1) ta n - 1 ? ( v 2 ) (2) 2 ta n - 1 ? ( 1 2 ) (3) ta n - 1 ? ( 1 2 ) (4) 2 ta n - 1 ? ( 1 v 5 ) Q4 - 2024 (29 Jan Shift 1) When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is ?? 8 , where x = Q5 - 2024 (29 Jan Shift 2) A simple harmonic oscillator has an amplitude A and time period 6 ?? second. Assuming the oscillation starts from its mean position, the time required by it to travel from ?? = ?? to ?? = v 3 2 ?? will be ?? x s, where x = Page 2 Previous Year Questions (2021-24): Simple Harmonic Motion (Oscillations) 2024 Q1 - 2024 (01 Feb Shift 2) A mass ?? is suspended from a spring of negligible mass and the system oscillates with a frequency ?? 1 . The frequency of oscillations if a mass 9 m is suspended from the same spring is f 2 . The value of f 1 f 2 is Q2 - 2024 (27 Jan Shift 1) A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is v ?? cm, where ?? = Q3 - 2024 (27 Jan Shift 2) A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ( ?? ) of thread deflection in the extreme position will be : (1) ta n - 1 ? ( v 2 ) (2) 2 ta n - 1 ? ( 1 2 ) (3) ta n - 1 ? ( 1 2 ) (4) 2 ta n - 1 ? ( 1 v 5 ) Q4 - 2024 (29 Jan Shift 1) When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is ?? 8 , where x = Q5 - 2024 (29 Jan Shift 2) A simple harmonic oscillator has an amplitude A and time period 6 ?? second. Assuming the oscillation starts from its mean position, the time required by it to travel from ?? = ?? to ?? = v 3 2 ?? will be ?? x s, where x = Questions with Answer Keys A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be s. [ take g = ?? 2 ms - 2 ] Q7 - 2024 (31 Jan Shift 1) A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2 A 3 . The new amplitude of motion is nA 3 . The value of n is Q8 - 2024 (31 Jan Shift 2) The time period of simple harmonic motion of mass ?? in the given figure is ?? v ???? 5 ?? , where the value of ?? is Answer Key Q1 (3) Q2 (12) Q3 (2) Q4 (9) Q5 (2) Q6 (8) Q7 (7) Q8 (12) Page 3 Previous Year Questions (2021-24): Simple Harmonic Motion (Oscillations) 2024 Q1 - 2024 (01 Feb Shift 2) A mass ?? is suspended from a spring of negligible mass and the system oscillates with a frequency ?? 1 . The frequency of oscillations if a mass 9 m is suspended from the same spring is f 2 . The value of f 1 f 2 is Q2 - 2024 (27 Jan Shift 1) A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is v ?? cm, where ?? = Q3 - 2024 (27 Jan Shift 2) A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ( ?? ) of thread deflection in the extreme position will be : (1) ta n - 1 ? ( v 2 ) (2) 2 ta n - 1 ? ( 1 2 ) (3) ta n - 1 ? ( 1 2 ) (4) 2 ta n - 1 ? ( 1 v 5 ) Q4 - 2024 (29 Jan Shift 1) When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is ?? 8 , where x = Q5 - 2024 (29 Jan Shift 2) A simple harmonic oscillator has an amplitude A and time period 6 ?? second. Assuming the oscillation starts from its mean position, the time required by it to travel from ?? = ?? to ?? = v 3 2 ?? will be ?? x s, where x = Questions with Answer Keys A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be s. [ take g = ?? 2 ms - 2 ] Q7 - 2024 (31 Jan Shift 1) A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2 A 3 . The new amplitude of motion is nA 3 . The value of n is Q8 - 2024 (31 Jan Shift 2) The time period of simple harmonic motion of mass ?? in the given figure is ?? v ???? 5 ?? , where the value of ?? is Answer Key Q1 (3) Q2 (12) Q3 (2) Q4 (9) Q5 (2) Q6 (8) Q7 (7) Q8 (12) Solutions Q1 f 1 = 1 2 ?? v k m f 2 = 1 2 ?? v k 9 m f 1 f 2 = v 9 1 = 3 1 Q2 V ut mean position = A ?? ? 10 = 4 ?? ?? = 5 2 v = ?? v A 2 - x 2 5 = 5 2 v 4 2 - x 2 ? x 2 = 16 - 4 x = v 12 cm Q3 Loss in kinetic energy = Gain in potential energy Page 4 Previous Year Questions (2021-24): Simple Harmonic Motion (Oscillations) 2024 Q1 - 2024 (01 Feb Shift 2) A mass ?? is suspended from a spring of negligible mass and the system oscillates with a frequency ?? 1 . The frequency of oscillations if a mass 9 m is suspended from the same spring is f 2 . The value of f 1 f 2 is Q2 - 2024 (27 Jan Shift 1) A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is v ?? cm, where ?? = Q3 - 2024 (27 Jan Shift 2) A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ( ?? ) of thread deflection in the extreme position will be : (1) ta n - 1 ? ( v 2 ) (2) 2 ta n - 1 ? ( 1 2 ) (3) ta n - 1 ? ( 1 2 ) (4) 2 ta n - 1 ? ( 1 v 5 ) Q4 - 2024 (29 Jan Shift 1) When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is ?? 8 , where x = Q5 - 2024 (29 Jan Shift 2) A simple harmonic oscillator has an amplitude A and time period 6 ?? second. Assuming the oscillation starts from its mean position, the time required by it to travel from ?? = ?? to ?? = v 3 2 ?? will be ?? x s, where x = Questions with Answer Keys A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be s. [ take g = ?? 2 ms - 2 ] Q7 - 2024 (31 Jan Shift 1) A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2 A 3 . The new amplitude of motion is nA 3 . The value of n is Q8 - 2024 (31 Jan Shift 2) The time period of simple harmonic motion of mass ?? in the given figure is ?? v ???? 5 ?? , where the value of ?? is Answer Key Q1 (3) Q2 (12) Q3 (2) Q4 (9) Q5 (2) Q6 (8) Q7 (7) Q8 (12) Solutions Q1 f 1 = 1 2 ?? v k m f 2 = 1 2 ?? v k 9 m f 1 f 2 = v 9 1 = 3 1 Q2 V ut mean position = A ?? ? 10 = 4 ?? ?? = 5 2 v = ?? v A 2 - x 2 5 = 5 2 v 4 2 - x 2 ? x 2 = 16 - 4 x = v 12 cm Q3 Loss in kinetic energy = Gain in potential energy ? 1 2 mv 2 = mg l ( 1 - co s ? ?? ) ? v 2 l = 2 g ( 1 - co s ? ?? ) Acceleration at lowest point = v 2 l Acceleration at extreme point = ?? s in ? ?? Hence, ?? 2 l = g s in ? ?? ? s in ? ?? = 2 ( 1 - co s ? ?? ) ? ta n ? ?? 2 = 1 2 ? ?? = 2 ta n - 1 ? ( 1 2 ) Q4 Let total energy = E = 1 2 KA 2 U = 1 2 K ( A 3 ) 2 = KA 2 2 × 9 = E 9 KE = E - E 9 = 8E 9 Ratio Total KE = E 8E 9 = 9 8 x = 9 Q5 From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of S H M. Page 5 Previous Year Questions (2021-24): Simple Harmonic Motion (Oscillations) 2024 Q1 - 2024 (01 Feb Shift 2) A mass ?? is suspended from a spring of negligible mass and the system oscillates with a frequency ?? 1 . The frequency of oscillations if a mass 9 m is suspended from the same spring is f 2 . The value of f 1 f 2 is Q2 - 2024 (27 Jan Shift 1) A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is v ?? cm, where ?? = Q3 - 2024 (27 Jan Shift 2) A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ( ?? ) of thread deflection in the extreme position will be : (1) ta n - 1 ? ( v 2 ) (2) 2 ta n - 1 ? ( 1 2 ) (3) ta n - 1 ? ( 1 2 ) (4) 2 ta n - 1 ? ( 1 v 5 ) Q4 - 2024 (29 Jan Shift 1) When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is ?? 8 , where x = Q5 - 2024 (29 Jan Shift 2) A simple harmonic oscillator has an amplitude A and time period 6 ?? second. Assuming the oscillation starts from its mean position, the time required by it to travel from ?? = ?? to ?? = v 3 2 ?? will be ?? x s, where x = Questions with Answer Keys A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be s. [ take g = ?? 2 ms - 2 ] Q7 - 2024 (31 Jan Shift 1) A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2 A 3 . The new amplitude of motion is nA 3 . The value of n is Q8 - 2024 (31 Jan Shift 2) The time period of simple harmonic motion of mass ?? in the given figure is ?? v ???? 5 ?? , where the value of ?? is Answer Key Q1 (3) Q2 (12) Q3 (2) Q4 (9) Q5 (2) Q6 (8) Q7 (7) Q8 (12) Solutions Q1 f 1 = 1 2 ?? v k m f 2 = 1 2 ?? v k 9 m f 1 f 2 = v 9 1 = 3 1 Q2 V ut mean position = A ?? ? 10 = 4 ?? ?? = 5 2 v = ?? v A 2 - x 2 5 = 5 2 v 4 2 - x 2 ? x 2 = 16 - 4 x = v 12 cm Q3 Loss in kinetic energy = Gain in potential energy ? 1 2 mv 2 = mg l ( 1 - co s ? ?? ) ? v 2 l = 2 g ( 1 - co s ? ?? ) Acceleration at lowest point = v 2 l Acceleration at extreme point = ?? s in ? ?? Hence, ?? 2 l = g s in ? ?? ? s in ? ?? = 2 ( 1 - co s ? ?? ) ? ta n ? ?? 2 = 1 2 ? ?? = 2 ta n - 1 ? ( 1 2 ) Q4 Let total energy = E = 1 2 KA 2 U = 1 2 K ( A 3 ) 2 = KA 2 2 × 9 = E 9 KE = E - E 9 = 8E 9 Ratio Total KE = E 8E 9 = 9 8 x = 9 Q5 From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of S H M. co s ? ?? = v 3 A 2 A = v 3 2 ?? = ?? 6 Now, ?? 6 = ?? t ?? 6 = 2 ?? ?? ?? ?? 6 = 2 ?? 6 ?? ?? t = ?? 2 So, ?? = 2 Q6 Acceleration due to gravity g ' = ?? 4 T = 2 ?? v 4 l g T = 2 ?? v 4 × 4 g T = 2 ?? 4 ?? = 8 s Q7 v = ?? v A 2 - x 2 at x = 2 A 3 ?? = ?? v ?? 2 - ( 2 ?? 3 ) 2 = v 5 A ?? 3 New amplitude = ?? 'Read More
289 videos|635 docs|179 tests
|
1. What is Simple Harmonic Motion (SHM) in oscillations? |
2. How is the period of oscillation related to the spring constant in Simple Harmonic Motion? |
3. What is the difference between frequency and angular frequency in the context of oscillations? |
4. How does the amplitude affect the energy of a Simple Harmonic Oscillator? |
5. Can the Simple Harmonic Motion be applied to real-life systems? |
|
Explore Courses for JEE exam
|