JEE Physics Oscillations MCQs Set 14

Physics Objective Questions and Answers: Oscillations

Review structured MCQ sets for JEE Physics Oscillations. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Oscillations MCQs with Answers

Access the complete set of multiple-choice questions for Oscillations below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A body executes SHM on a straight line ABC, with its extreme positions at A and C. The distances are AB = a and BC = b. The speed of the particle at the mid-point of the line AC is u. The time period of the SHM is:
(a) \( \frac{\pi(a+b)}{u} \)
(b) \( \frac{2\pi(b-a)}{u} \)
(c) \( \frac{\pi(b-a)}{u} \)
(d) \( \frac{2\pi(a+b)}{u} \)
Show Answer & Explanation

Answer: (a) \( \frac{\pi(a+b)}{u} \)

Explanation:
1. The amplitude is half the total length of the path: \( A=\frac{a+b}{2} \).
2. The speed at the mid-point (mean position) is the maximum speed: \( u=\omega A=\frac{2\pi}{T}\cdot\frac{a+b}{2} \), so \( T=\frac{\pi(a+b)}{u} \).

Teacher's Note:
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77. .

Question 2: A spring has an equilibrium length of 2.0 m and a spring constant of 10 N/m. Alice pulls on one end with a force of 3.0 N and Bob pulls on the opposite end with a force of 3.0 N in the opposite direction. The resulting length of the spring is:
(a) 1.7 m
(b) 2.3 m
(c) 2.0 m
(d) 2.6 m
Show Answer & Explanation

Answer: (b) 2.3 m

Explanation:
1. The tension in the spring is 3 N (not 6 N), because each end is balanced by the other.
2. Extension \( =\frac{3}{10}=0.3 \) m, so the length is 2.3 m.

Teacher's Note:
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80. .

Question 3: An object of mass 0.2 kg executes simple harmonic oscillations along the X-axis with frequency \( \frac{25}{\pi} \) Hz. At the position x = 0.04 m, the object has kinetic energy 0.5 J and potential energy 0.4 J (take the potential energy as zero at the equilibrium position). The amplitude of the oscillations is:
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 5 cm
Show Answer & Explanation

Answer: (c) 6 cm

Explanation:
1. \( \omega=2\pi f=50 \) rad/s. Total energy \( =0.5+0.4=0.9 \) J.
2. \( E=\frac12m\omega^2A^2=\frac12(0.2)(2500)A^2=250A^2=0.9 \), so \( A=0.06 \) m \( =6 \) cm.

Teacher's Note:
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117. .

Question 4: The displacement of a particle varies with time according to \( y=a\sin\omega t+b\cos\omega t \). The motion is:
(a) oscillatory but not SHM
(b) SHM with amplitude \( a+b \)
(c) SHM with amplitude \( a^2+b^2 \)
(d) SHM with amplitude \( \sqrt{a^2+b^2} \)
Show Answer & Explanation

Answer: (d) SHM with amplitude \( \sqrt{a^2+b^2} \)

Explanation:
1. Write \( y=R\sin(\omega t+\phi) \) with \( R\cos\phi=a \) and \( R\sin\phi=b \).
2. Then \( R=\sqrt{a^2+b^2} \): the motion is SHM with this amplitude.

Teacher's Note:
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72. .

Question 5: A mass m is connected individually to two springs \( S_1 \) and \( S_2 \) and the oscillation frequencies are \( \nu_1 \) and \( \nu_2 \). If the same mass is attached between the two springs (one on each side, on a smooth horizontal surface), the oscillation frequency would be:
(a) \( \sqrt{\nu_1^2+\nu_2^2} \)
(b) \( \nu_1+\nu_2 \)
(c) \( \left(\frac1{\nu_1}+\frac1{\nu_2}\right)^{-1} \)
(d) \( \sqrt{\nu_1^2-\nu_2^2} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\nu_1^2+\nu_2^2} \)

Explanation:
1. Springs on the two sides of the mass act in parallel: \( k=k_1+k_2 \).
2. \( \nu=\frac1{2\pi}\sqrt{\frac{k_1+k_2}{m}} \), so \( \nu^2=\nu_1^2+\nu_2^2 \).

Teacher's Note:
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107. .

Question 6: A particle executing simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude is:
(a) 1/4 s
(b) 1/2 s
(c) 3/4 s
(d) 3/2 s
Show Answer & Explanation

Answer: (b) 1/2 s

Explanation:
1. Starting from the mean position, \( x=A\sin\frac{2\pi t}{T} \).
2. \( \frac A2=A\sin\frac{2\pi t}{6} \) gives \( \frac{2\pi t}{6}=\frac\pi6 \), so \( t=\frac12 \) s.

Teacher's Note:
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102. .

Question 7: The total energy of a particle executing simple harmonic motion depends on its: (I) amplitude, (II) period, (III) displacement. Which are correct?
(a) II and III only
(b) I and III only
(c) I and II only
(d) I, II and III
Show Answer & Explanation

Answer: (c) I and II only

Explanation:
1. \( E=\frac12m\omega^2A^2=\frac{2\pi^2mA^2}{T^2} \): it depends on the amplitude and on the period.
2. It does not depend on the instantaneous displacement, because kinetic and potential energies exchange in a constant total.

Teacher's Note:
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Question 8: A mass m is attached to four springs on a frictionless horizontal floor: on the left two springs of constants 2k and 2k in series, and on the right two springs of constant k each, in parallel (both attached to the mass and to the wall). The frequency of the resulting SHM for a small displacement is:
(a) \( \frac1{2\pi}\sqrt{\frac{11k}{2m}} \)
(b) \( \frac1{2\pi}\sqrt{\frac{2k}{3m}} \)
(c) \( \frac1{2\pi}\sqrt{\frac{4k}{m}} \)
(d) \( \frac1{2\pi}\sqrt{\frac{3k}{m}} \)
Show Answer & Explanation

Answer: (d) \( \frac1{2\pi}\sqrt{\frac{3k}{m}} \)

Explanation:
1. Left side: \( 2k \) and \( 2k \) in series give \( \frac{2k\cdot2k}{4k}=k \). Right side: \( k+k=2k \).
2. Both sides act on the mass in parallel: \( k_{eff}=k+2k=3k \), so \( f=\frac1{2\pi}\sqrt{\frac{3k}{m}} \).

Teacher's Note:
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121. .

Question 9: A flat plate P of mass M slides on a frictionless horizontal surface executing SHM with frequency \( \nu \). A block B of mass m rests on the plate, with coefficient of friction \( \mu_s \) between B and P. If the block is not to slip, the maximum amplitude of oscillation of the plate-block system is:
(a) \( \frac{\mu_sg}{4\pi^2\nu^2} \)
(b) \( \frac{\mu_sg}{4\pi\nu^2} \)
(c) \( \frac{\mu_s^2g}{\pi^2\nu^2} \)
(d) \( \frac{\mu_s^2g}{4\pi^2\nu^2} \)
Show Answer & Explanation

Answer: (a) \( \frac{\mu_sg}{4\pi^2\nu^2} \)

Explanation:
1. The block's acceleration is provided by friction: \( m\omega^2A\le\mu_smg \).
2. \( A_{max}=\frac{\mu_sg}{\omega^2}=\frac{\mu_sg}{4\pi^2\nu^2} \).

Teacher's Note:
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117. .

Question 10: The ratio of the kinetic energy at the mean position to the potential energy at a displacement A/2 of a particle performing SHM of amplitude A is:
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1
Show Answer & Explanation

Answer: (b) 4 : 1

Explanation:
1. KE at the mean position \( =\frac12m\omega^2A^2 \).
2. PE at \( \frac A2 \) \( =\frac12m\omega^2\frac{A^2}{4} \). Ratio \( =4:1 \).

Teacher's Note:
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81. .

Question 11: The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5 s. In another 10 s it will decrease to \( \alpha \) times its original magnitude, where \( \alpha \) equals:
(a) 0.7
(b) 0.81
(c) 0.729
(d) 0.6
Show Answer & Explanation

Answer: (c) 0.729

Explanation:
1. The amplitude decays exponentially, so it multiplies by the same factor in equal time intervals: 0.9 for every 5 s.
2. After another 10 s (two more intervals): \( \alpha=0.9\times0.9\times0.9=0.729 \).

Teacher's Note:
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63. .

Question 12: If a simple pendulum has significant amplitude (up to a factor of 1/e of the original) only in the period between t = 0 and \( t=\tau \), then \( \tau \) may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with b as the constant of proportionality, the average life time of the pendulum (assuming damping is small) in seconds is:
(a) \( \frac{0.693}{b} \)
(b) \( b \)
(c) \( \frac1b \)
(d) \( \frac2b \)
Show Answer & Explanation

Answer: (d) \( \frac2b \)

Explanation:
1. Retardation \( =bv \) gives the equation \( \ddot x+b\dot x+\omega^2x=0 \), whose solution has amplitude \( A_0e^{-bt/2} \).
2. The amplitude falls to \( \frac1e \) when \( \frac{b\tau}{2}=1 \), so \( \tau=\frac2b \).

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70. s
71.
72. t
73. i
74. m
75. e
76.
77. c
78. o
79. n
80. s
81. t
82. a
83. n
84. t
85.
86. i
87. s
88.
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90. w
91. i
92. c
93. e
94.
95. a
96. s
97.
98. l
99. o
100. n
101. g
102. .

Question 13: For a particle executing simple harmonic motion, the maximum acceleration is \( \alpha \) and the maximum velocity is \( \beta \). Its time period of vibration is:
(a) \( \frac{2\pi\beta}{\alpha} \)
(b) \( \frac{\beta^2}{\alpha} \)
(c) \( \frac{\beta^2}{\alpha^2} \)
(d) \( \frac\alpha\beta \)
Show Answer & Explanation

Answer: (a) \( \frac{2\pi\beta}{\alpha} \)

Explanation:
1. \( \alpha=\omega^2A \) and \( \beta=\omega A \), so \( \omega=\frac\alpha\beta \).
2. \( T=\frac{2\pi}{\omega}=\frac{2\pi\beta}{\alpha} \).

Teacher's Note:
1. T
2. h
3. e
4.
5. r
6. a
7. t
8. i
9. o
10.
11. o
12. f
13.
14. m
15. a
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17. i
18. m
19. u
20. m
21.
22. a
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24. c
25. e
26. l
27. e
28. r
29. a
30. t
31. i
32. o
33. n
34.
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36. o
37.
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40. x
41. i
42. m
43. u
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45.
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47. e
48. l
49. o
50. c
51. i
52. t
53. y
54.
55. g
56. i
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58. e
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60.
61. t
62. h
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64.
65. a
66. n
67. g
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69. l
70. a
71. r
72.
73. f
74. r
75. e
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77. u
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79. n
80. c
81. y
82.
83. d
84. i
85. r
86. e
87. c
88. t
89. l
90. y
91. .

Question 14: A particle is executing SHM along a straight line. Its velocities at distances \( x_1 \) and \( x_2 \) from the mean position are \( V_1 \) and \( V_2 \) respectively. Its time period is:
(a) \( 2\pi\sqrt{\frac{V_1^2+V_2^2}{x_1^2+x_2^2}} \)
(b) \( 2\pi\sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}} \)
(c) \( 2\pi\sqrt{\frac{V_1^2-V_2^2}{x_1^2-x_2^2}} \)
(d) \( 2\pi\sqrt{\frac{x_1^2+x_2^2}{V_1^2+V_2^2}} \)
Show Answer & Explanation

Answer: (b) \( 2\pi\sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}} \)

Explanation:
1. \( V^2=\omega^2(A^2-x^2) \). Subtract for the two positions: \( V_1^2-V_2^2=\omega^2(x_2^2-x_1^2) \).
2. \( \omega=\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}} \), so \( T=2\pi\sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}} \).

Teacher's Note:
1. S
2. u
3. b
4. t
5. r
6. a
7. c
8. t
9. i
10. n
11. g
12.
13. t
14. h
15. e
16.
17. t
18. w
19. o
20.
21. v
22. e
23. l
24. o
25. c
26. i
27. t
28. y
29. -
30. d
31. i
32. s
33. p
34. l
35. a
36. c
37. e
38. m
39. e
40. n
41. t
42.
43. r
44. e
45. l
46. a
47. t
48. i
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50. n
51. s
52.
53. e
54. l
55. i
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57. i
58. n
59. a
60. t
61. e
62. s
63.
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65. h
66. e
67.
68. a
69. m
70. p
71. l
72. i
73. t
74. u
75. d
76. e
77. .

Question 15: Two displacements \( y_1=a\sin\omega t \) and \( y_2=b\cos\omega t \) are superimposed. The resulting motion is:
(a) simple harmonic with amplitude \( \frac{a^2+b^2}{2} \)
(b) not simple harmonic
(c) simple harmonic with amplitude \( \sqrt{a^2+b^2} \)
(d) simple harmonic with amplitude \( \frac ab \)
Show Answer & Explanation

Answer: (c) simple harmonic with amplitude \( \sqrt{a^2+b^2} \)

Explanation:
1. The two are in quadrature (phase difference \( \frac\pi2 \)) and have the same frequency.
2. The resultant is \( \sqrt{a^2+b^2}\sin(\omega t+\phi) \): SHM with amplitude \( \sqrt{a^2+b^2} \).

Teacher's Note:
1. P
2. e
3. r
4. p
5. e
6. n
7. d
8. i
9. c
10. u
11. l
12. a
13. r
14.
15. p
16. h
17. a
18. s
19. o
20. r
21. s
22.
23. o
24. f
25.
26. l
27. e
28. n
29. g
30. t
31. h
32. s
33.
34. a
35.
36. a
37. n
38. d
39.
40. b
41.
42. a
43. d
44. d
45.
46. b
47. y
48.
49. P
50. y
51. t
52. h
53. a
54. g
55. o
56. r
57. a
58. s
59. .

Question 16: Which of the following functions represent simple harmonic motion? (I) \( y=\sin\omega t-\cos\omega t \). (II) \( y=\sin^3\omega t \). (III) \( y=5\cos\left(\frac{3\pi}{4}-3\omega t\right) \). (IV) \( y=1+\omega t+\omega^2t^2 \).
(a) I only
(b) II and IV only
(c) I and II only
(d) I and III only
Show Answer & Explanation

Answer: (d) I and III only

Explanation:
1. (I) \( =\sqrt2\sin(\omega t-\frac\pi4) \): SHM. (III) is a cosine of a linear function of time: SHM with angular frequency \( 3\omega \).
2. (II) \( \sin^3\omega t=\frac{3\sin\omega t-\sin3\omega t}{4} \) is periodic but a sum of two harmonics, not SHM. (IV) is not even periodic.

Teacher's Note:
1. S
2. H
3. M
4.
5. i
6. s
7.
8. a
9.
10. s
11. i
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20. u
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25. :
26.
27. p
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35.
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37. u
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71.
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74. e
75.
76. n
77. o
78. t
79.
80. S
81. H
82. M
83. .

Question 17: Two particles oscillate along two close parallel straight lines side by side with the same frequency and amplitude. They pass each other, moving in opposite directions, when their displacement is half of the amplitude. The phase difference between them is:
(a) \( \frac{2\pi}{3} \)
(b) \( \frac\pi6 \)
(c) 0
(d) \( \pi \)
Show Answer & Explanation

Answer: (a) \( \frac{2\pi}{3} \)

Explanation:
1. At displacement \( \frac A2 \) the phase is \( \frac\pi6 \) (moving outward) or \( \frac{5\pi}{6} \) (moving back).
2. Opposite directions means the phases are \( \frac\pi6 \) and \( \frac{5\pi}{6} \), so the phase difference is \( \frac{4\pi}{6}=\frac{2\pi}{3} \).

Teacher's Note:
1. F
2. o
3. r
4.
5. a
6.
7. g
8. i
9. v
10. e
11. n
12.
13. d
14. i
15. s
16. p
17. l
18. a
19. c
20. e
21. m
22. e
23. n
24. t
25.
26. t
27. h
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30. e
31.
32. a
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42. a
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75. o
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77. i
78. o
79. n
80. .

Question 18: The displacement of a particle along the x-axis is given by \( x=a\sin^2\omega t \). The motion corresponds to:
(a) simple harmonic motion of frequency \( \frac{3\omega}{2\pi} \)
(b) simple harmonic motion of frequency \( \frac\omega\pi \)
(c) non-simple harmonic motion
(d) simple harmonic motion of frequency \( \frac{\omega}{2\pi} \)
Show Answer & Explanation

Answer: (b) simple harmonic motion of frequency \( \frac\omega\pi \)

Explanation:
1. \( x=a\sin^2\omega t=\frac a2(1-\cos2\omega t) \).
2. This is SHM about the point \( x=\frac a2 \) with angular frequency \( 2\omega \), that is, frequency \( \frac{2\omega}{2\pi}=\frac\omega\pi \).

Teacher's Note:
1. U
2. s
3. e
4.
5. a
6.
7. d
8. o
9. u
10. b
11. l
12. e
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92. .

Question 19: A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground. The masses of the spring and pan are negligible. When pressed slightly and released, the mass executes simple harmonic motion. The spring constant is 200 N/m. The minimum amplitude of the motion for which the mass just gets detached from the pan is (g = 10 m/s²):
(a) 4.0 cm
(b) 8.0 cm
(c) 10.0 cm
(d) any value less than 12.0 cm
Show Answer & Explanation

Answer: (c) 10.0 cm

Explanation:
1. The mass leaves the pan when the pan's downward acceleration at the top equals g: \( \omega^2A=g \).
2. \( \omega^2=\frac{k}{m}=100 \), so \( A=\frac{g}{\omega^2}=0.10 \) m \( =10 \) cm (detachment occurs for amplitudes above this).

Teacher's Note:
1. C
2. o
3. n
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5. a
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7. t
8.
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11.
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65. p
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72.
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116. e
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118.
119. g
120. .

Question 20: A particle executing simple harmonic motion has kinetic energy \( K_0\cos^2\omega t \). The maximum values of the potential energy and the total energy are respectively:
(a) \( \frac{K_0}{2}\ \text{and}\ K_0 \)
(b) \( K_0\ \text{and}\ 2K_0 \)
(c) \( 0\ \text{and}\ 2K_0 \)
(d) \( K_0\ \text{and}\ K_0 \)
Show Answer & Explanation

Answer: (d) \( K_0\ \text{and}\ K_0 \)

Explanation:
1. The maximum kinetic energy is \( K_0 \), and this equals the total energy.
2. The potential energy is \( E-K=K_0\sin^2\omega t \), so its maximum is also \( K_0 \).

Teacher's Note:
1. I
2. n
3.
4. S
5. H
6. M
7.
8. t
9. h
10. e
11.
12. t
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33. o
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42. a
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85. t
86. i
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89.
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91. n
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94. g
95. y
96. .

Question 21: The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, the new time period will be:
(a) \( \frac T2 \)
(b) \( \frac T4 \)
(c) T
(d) 2T
Show Answer & Explanation

Answer: (a) \( \frac T2 \)

Explanation:
1. Each quarter has four times the spring constant: \( k'=4k \).
2. \( T'=2\pi\sqrt{\frac{m}{4k}}=\frac T2 \).

Teacher's Note:
1. S
2. t
3. i
4. f
5. f
6. n
7. e
8. s
9. s
10.
11. i
12. s
13.
14. i
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19. s
20. e
21. l
22. y
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24. p
25. r
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27. p
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29. r
30. t
31. i
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34. a
35. l
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37. t
38. o
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43.
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61. g
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81.
82. a
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84. e
85.
86. s
87. t
88. i
89. f
90. f
91. e
92. r
93. .

Question 22: A mass is suspended separately from two different springs and has time periods \( t_1 \) and \( t_2 \) respectively. If it is connected to both springs side by side (in parallel), as shown, the time period \( t_0 \) is given by:
(a) \( t_0^2=t_1^2+t_2^2 \)
(b) \( t_0^{-2}=t_1^{-2}+t_2^{-2} \)
(c) \( t_0^{-1}=t_1^{-1}+t_2^{-1} \)
(d) \( t_0=t_1+t_2 \)
Show Answer & Explanation

Answer: (b) \( t_0^{-2}=t_1^{-2}+t_2^{-2} \)

Explanation:
1. Parallel springs: \( k=k_1+k_2 \), and \( k\propto\frac1{T^2} \).
2. So \( \frac1{t_0^2}=\frac1{t_1^2}+\frac1{t_2^2} \).

Teacher's Note:
1. S
2. i
3. n
4. c
5. e
6.
7. \
8. (
9.
10. k
11. =
12. \
13. f
14. r
15. a
16. c
17. {
18. 4
19. \
20. p
21. i
22. ^
23. 2
24. m
25. }
26. {
27. T
28. ^
29. 2
30. }
31.
32. \
33. )
34. ,
35.
36. s
37. p
38. r
39. i
40. n
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42. s
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44. i
45. n
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52. l
53. e
54. l
55.
56. a
57. d
58. d
59.
60. a
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62.
63. i
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65. v
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69. e
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71. s
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73. u
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81.
82. t
83. h
84. e
85.
86. p
87. e
88. r
89. i
90. o
91. d
92. s
93. .

Question 23: A particle with a restoring force proportional to displacement and a resisting force proportional to velocity is subjected to a force \( F\sin\omega_0t \) (forced damped oscillation, where \( \omega_0 \) is the driving frequency and the natural frequency of the undamped oscillator is \( \omega_n \)). If the amplitude is maximum at \( \omega_0=\omega_1 \) and the kinetic energy of the particle (velocity amplitude) is maximum at \( \omega_0=\omega_2 \), then:
(a) \( \omega_1=\omega_n\ \text{and}\ \omega_2=\omega_n \)
(b) \( \omega_1=\omega_n\ \text{and}\ \omega_2\ne\omega_n \)
(c) \( \omega_1\ne\omega_n\ \text{and}\ \omega_2=\omega_n \)
(d) \( \omega_1\ne\omega_n\ \text{and}\ \omega_2\ne\omega_n \)
Show Answer & Explanation

Answer: (c) \( \omega_1\ne\omega_n\ \text{and}\ \omega_2=\omega_n \)

Explanation:
1. With damping, the displacement amplitude is maximum at \( \omega_1=\sqrt{\omega_n^2-\frac{b^2}{2m^2}} \), which is slightly less than \( \omega_n \).
2. The velocity amplitude (and therefore the kinetic energy) is maximum exactly at \( \omega_2=\omega_n \).

Teacher's Note:
1. D
2. i
3. s
4. p
5. l
6. a
7. c
8. e
9. m
10. e
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20. n
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24. i
25. s
26.
27. s
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30. g
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34. y
35.
36. b
37. e
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61. c
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93. a
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96. l
97. y
98.
99. a
100. t
101.
102. i
103. t
104. .

Question 24: Two SHMs of the same amplitude and time period act together in perpendicular directions with a phase difference of \( \frac\pi2 \). They give rise to:
(a) straight line motion
(b) elliptical motion which is not circular
(c) none of these
(d) circular motion
Show Answer & Explanation

Answer: (d) circular motion

Explanation:
1. \( x=A\sin\omega t \) and \( y=A\cos\omega t \).
2. Then \( x^2+y^2=A^2 \): the particle moves in a circle.

Teacher's Note:
1. W
2. i
3. t
4. h
5.
6. u
7. n
8. e
9. q
10. u
11. a
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13.
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39.
40. d
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FAQs

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