Exams Net
Unrestricted Exams Practice
Home
Exams
Banking
CUET
Defence
Engineering
Finance
GATE
Insurance
International
JEE
LAW
MBA
MCA
Medical
Other
Police
PSC
RRB
SSC
State Govt
Subjectwise
Teacher
SET Exams
UPSC
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
CBSE
CBSE Class 10 Solutions
CBSE Class 12 Solutions
CBSE Question Papers (Pdf)
NCERT Books (Pdf)
NCERT Exemplar Books (Pdf)
NCERT Study Notes (Pdf)
CBSE Study Concepts (Pdf)
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Contact Us
Our Apps
Privacy
+
-
Test Index
Oscillations Part 2
Show Para
Hide Para
Section:
Physics
Share question:
© examsnet.com
Question : 15 of 81
Marks:
+1
,
-0
A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is
200
N/m
200 \text{N/m}
200
N/m
. The block is pushed such that the length of the spring becomes 1 m and then released. At distance
x
m
(
x
<
2
)
x \text{m} (x < 2)
x
m
(
x
<
2
)
from the wall, the speed of the block will be
[8 Apr 2025 Shift 2]
10
[
1
−
(
2
−
x
)
2
]
2
m/s
10 \left[ 1-(2-x)^2 \right]^2 \text{m/s}
10
[
1
−
(
2
−
x
)
2
]
2
m/s
10
[
1
−
(
2
−
x
)
2
]
1
2
m/s
10 \left[ 1-(2-x)^2 \right]^{\frac{1}{2}} \text{m/s}
10
[
1
−
(
2
−
x
)
2
]
2
1
​
m/s
10
[
1
−
(
2
−
x
)
]
3
2
m/s
10 [1-(2-x)]^{\frac{3}{2}} \text{m/s}
10
[
1
−
(
2
−
x
)
]
2
3
​
m/s
10
[
1
−
(
2
−
x
)
2
]
m/s
10 \left[ 1-(2-x)^2 \right] \text{m/s}
10
[
1
−
(
2
−
x
)
2
]
m/s
Validate
Solution:
Energy conservation
  
1
2
k
(
1
)
2
=
  
1
2
m
v
2
+
  
1
2
k
(
2
−
x
)
2
\; \frac{1}{2} k(1)^2 = \; \frac{1}{2} m v^2 + \; \frac{1}{2} k(2-x)^2
2
1
​
k
(
1
)
2
=
2
1
​
m
v
2
+
2
1
​
k
(
2
−
x
)
2
Compression in the spring
=
(
2
−
x
)
=(2-x)
=
(
2
−
x
)
⇒
  
1
2
k
(
1
−
(
2
−
x
)
2
)
=
  
1
2
m
v
2
\Rightarrow \; \frac{1}{2} k (1-(2-x)^2) = \; \frac{1}{2} m v^2
⇒
2
1
​
k
(
1
−
(
2
−
x
)
2
)
=
2
1
​
m
v
2
⇒
v
=
[
100
(
1
−
(
2
−
x
)
2
)
]
1
2
\Rightarrow v = \left[ 100 (1-(2-x)^2) \right]^{\frac{1}{2}}
⇒
v
=
[
100
(
1
−
(
2
−
x
)
2
)
]
2
1
​
  
v
=
10
[
1
−
(
2
−
x
)
]
1
2
\; v = 10 [1-(2-x)]^{\frac{1}{2}}
v
=
10
[
1
−
(
2
−
x
)
]
2
1
​
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
Prev Question
Next Question
More Free Exams
AIEEE Previous Papers
BITSAT Exam Previous Papers
JEE Adv
JEE Advanced Model Papers
JEE Advanced Previous Papers
JEE Main PYQ
JEE Mains Model Papers
VITEEE Previous Papers
