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
(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:
1. T
2. h
3. e
4.
5. m
6. a
7. x
8. i
9. m
10. u
11. m
12.
13. s
14. p
15. e
16. e
17. d
18.
19. o
20. f
21.
22. a
23. n
24.
25. S
26. H
27. M
28.
29. e
30. q
31. u
32. a
33. l
34. s
35.
36. t
37. h
38. e
39.
40. a
41. m
42. p
43. l
44. i
45. t
46. u
47. d
48. e
49.
50. t
51. i
52. m
53. e
54. s
55.
56. t
57. h
58. e
59.
60. a
61. n
62. g
63. u
64. l
65. a
66. r
67.
68. f
69. r
70. e
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72. u
73. e
74. n
75. c
76. y
77. .
(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:
1. A
2.
3. s
4. p
5. r
6. i
7. n
8. g
9.
10. p
11. u
12. l
13. l
14. e
15. d
16.
17. e
18. q
19. u
20. a
21. l
22. l
23. y
24.
25. f
26. r
27. o
28. m
29.
30. b
31. o
32. t
33. h
34.
35. e
36. n
37. d
38. s
39.
40. h
41. a
42. s
43.
44. a
45.
46. t
47. e
48. n
49. s
50. i
51. o
52. n
53.
54. e
55. q
56. u
57. a
58. l
59.
60. t
61. o
62.
63. o
64. n
65. e
66.
67. o
68. f
69.
70. t
71. h
72. e
73.
74. f
75. o
76. r
77. c
78. e
79. s
80. .
(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:
1. T
2. h
3. e
4.
5. t
6. o
7. t
8. a
9. l
10.
11. e
12. n
13. e
14. r
15. g
16. y
17.
18. o
19. f
20.
21. a
22. n
23.
24. S
25. H
26. M
27.
28. i
29. s
30.
31. t
32. h
33. e
34.
35. s
36. u
37. m
38.
39. o
40. f
41.
42. t
43. h
44. e
45.
46. k
47. i
48. n
49. e
50. t
51. i
52. c
53.
54. a
55. n
56. d
57.
58. p
59. o
60. t
61. e
62. n
63. t
64. i
65. a
66. l
67.
68. e
69. n
70. e
71. r
72. g
73. y
74.
75. a
76. t
77.
78. a
79. n
80. y
81.
82. i
83. n
84. s
85. t
86. a
87. n
88. t
89. ,
90.
91. a
92. n
93. d
94.
95. i
96. t
97.
98. f
99. i
100. x
101. e
102. s
103.
104. t
105. h
106. e
107.
108. a
109. m
110. p
111. l
112. i
113. t
114. u
115. d
116. e
117. .
(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:
1. A
2.
3. s
4. u
5. m
6.
7. o
8. f
9.
10. a
11.
12. s
13. i
14. n
15. e
16.
17. a
18. n
19. d
20.
21. a
22.
23. c
24. o
25. s
26. i
27. n
28. e
29.
30. o
31. f
32.
33. t
34. h
35. e
36.
37. s
38. a
39. m
40. e
41.
42. f
43. r
44. e
45. q
46. u
47. e
48. n
49. c
50. y
51.
52. i
53. s
54.
55. a
56.
57. s
58. i
59. n
60. g
61. l
62. e
63.
64. s
65. i
66. n
67. u
68. s
69. o
70. i
71. d
72. .
(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:
1. F
2. r
3. e
4. q
5. u
6. e
7. n
8. c
9. y
10.
11. s
12. q
13. u
14. a
15. r
16. e
17. d
18.
19. i
20. s
21.
22. p
23. r
24. o
25. p
26. o
27. r
28. t
29. i
30. o
31. n
32. a
33. l
34.
35. t
36. o
37.
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39. h
40. e
41.
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43. p
44. r
45. i
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47. g
48.
49. c
50. o
51. n
52. s
53. t
54. a
55. n
56. t
57. ,
58.
59. s
60. o
61.
62. e
63. f
64. f
65. e
66. c
67. t
68. i
69. v
70. e
71.
72. s
73. p
74. r
75. i
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77. g
78. s
79.
80. a
81. d
82. d
83.
84. i
85. n
86.
87. s
88. q
89. u
90. a
91. r
92. e
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94.
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96. f
97.
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100. e
101. q
102. u
103. e
104. n
105. c
106. y
107. .
(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:
1. F
2. r
3. o
4. m
5.
6. t
7. h
8. e
9.
10. m
11. e
12. a
13. n
14.
15. p
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20. i
21. o
22. n
23.
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25. o
26.
27. h
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30. f
31.
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34. p
35. l
36. i
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38. u
39. d
40. e
41.
42. t
43. a
44. k
45. e
46. s
47.
48. T
49. /
50. 1
51. 2
52. ,
53.
54. a
55. n
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57.
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59. r
60. o
61. m
62.
63. h
64. a
65. l
66. f
67.
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69. m
70. p
71. l
72. i
73. t
74. u
75. d
76. e
77.
78. t
79. o
80.
81. t
82. h
83. e
84.
85. e
86. x
87. t
88. r
89. e
90. m
91. e
92.
93. t
94. a
95. k
96. e
97. s
98.
99. T
100. /
101. 6
102. .
(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:
1. T
2. h
3. e
4.
5. t
6. o
7. t
8. a
9. l
10.
11. e
12. n
13. e
14. r
15. g
16. y
17.
18. i
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20.
21. c
22. o
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25. t
26. a
27. n
28. t
29.
30. d
31. u
32. r
33. i
34. n
35. g
36.
37. t
38. h
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40.
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42. o
43. t
44. i
45. o
46. n
47. ,
48.
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50. o
51.
52. i
53. t
54.
55. c
56. a
57. n
58. n
59. o
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61.
62. d
63. e
64. p
65. e
66. n
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68.
69. o
70. n
71.
72. t
73. h
74. e
75.
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77. i
78. s
79. p
80. l
81. a
82. c
83. e
84. m
85. e
86. n
87. t
88. .
(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:
1. R
2. e
3. d
4. u
5. c
6. e
7.
8. e
9. a
10. c
11. h
12.
13. s
14. i
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31. e
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33. t
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43. i
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46. t
47. ,
48.
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50. n
51. d
52.
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54. h
55. e
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58. c
59. o
60. m
61. b
62. i
63. n
64. e
65.
66. t
67. h
68. e
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70. t
71. w
72. o
73.
74. s
75. i
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77. e
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79.
80. a
81. s
82.
83. s
84. p
85. r
86. i
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90.
91. a
92. c
93. t
94. i
95. n
96. g
97.
98. i
99. n
100.
101. p
102. a
103. r
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106. l
107. e
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109.
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114. h
115. e
116.
117. m
118. a
119. s
120. s
121. .
(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:
1. T
2. h
3. e
4.
5. m
6. a
7. x
8. i
9. m
10. u
11. m
12.
13. a
14. c
15. c
16. e
17. l
18. e
19. r
20. a
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24. n
25.
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27. f
28.
29. t
30. h
31. e
32.
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34. y
35. s
36. t
37. e
38. m
39.
40. i
41. s
42.
43. \
44. (
45.
46. \
47. o
48. m
49. e
50. g
51. a
52. ^
53. 2
54. A
55.
56. \
57. )
58. ,
59.
60. a
61. n
62. d
63.
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65. r
66. i
67. c
68. t
69. i
70. o
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72.
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77. h
78. e
79.
80. b
81. l
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91. u
92. p
93. p
94. l
95. y
96.
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101. o
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103. t
104.
105. \
106. (
107.
108. \
109. m
110. u
111. _
112. s
113. g
114.
115. \
116. )
117. .
(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:
1. P
2. o
3. t
4. e
5. n
6. t
7. i
8. a
9. l
10.
11. e
12. n
13. e
14. r
15. g
16. y
17.
18. s
19. c
20. a
21. l
22. e
23. s
24.
25. a
26. s
27.
28. t
29. h
30. e
31.
32. s
33. q
34. u
35. a
36. r
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38.
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41.
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43. h
44. e
45.
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47. i
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50. l
51. a
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53. e
54. m
55. e
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58.
59. f
60. r
61. o
62. m
63.
64. t
65. h
66. e
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68. m
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70. a
71. n
72.
73. p
74. o
75. s
76. i
77. t
78. i
79. o
80. n
81. .
(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:
1. E
2. x
3. p
4. o
5. n
6. e
7. n
8. t
9. i
10. a
11. l
12.
13. d
14. e
15. c
16. a
17. y
18.
19. m
20. e
21. a
22. n
23. s
24.
25. e
26. q
27. u
28. a
29. l
30.
31. t
32. i
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34. e
35.
36. i
37. n
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39. e
40. r
41. v
42. a
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45.
46. g
47. i
48. v
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50.
51. e
52. q
53. u
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55. l
56.
57. r
58. a
59. t
60. i
61. o
62. s
63. .
(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 \( \tau=\frac2b \).
Teacher's Note:
1. T
2. h
3. e
4.
5. a
6. m
7. p
8. l
9. i
10. t
11. u
12. d
13. e
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21.
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25. h
26. a
27. l
28. f
29.
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31. h
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33.
34. r
35. a
36. t
37. e
38.
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40. t
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42. w
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49. h
50. e
51.
52. e
53. n
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57. y
58.
59. d
60. o
61. e
62. s
63. ,
64.
65. s
66. o
67.
68. i
69. t
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.
89. t
90. w
91. i
92. c
93. e
94.
95. a
96. s
97.
98. l
99. o
100. n
101. g
102. .
(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
16. x
17. i
18. m
19. u
20. m
21.
22. a
23. c
24. c
25. e
26. l
27. e
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33. n
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56. i
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83. d
84. i
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91. .
(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 \( T=2\pi\sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}} \).
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(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 \( \sqrt{a^2+b^2} \).
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(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.
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83. .
(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 \( \frac{4\pi}{6}=\frac{2\pi}{3} \).
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80. .
(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 \( \frac{2\omega}{2\pi}=\frac\omega\pi \).
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92. .
(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).
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120. .
(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 \( K_0 \).
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96. .
(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. \( T'=2\pi\sqrt{\frac{m}{4k}}=\frac T2 \).
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93. .
(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 \( \frac1{t_0^2}=\frac1{t_1^2}+\frac1{t_2^2} \).
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93. .
(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 \( \omega_2=\omega_n \).
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104. .
(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.
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