Solve
Guides
Join / Login
Use app
Login
0
You visited us
0
times! Enjoying our articles?
Unlock Full Access!
Standard XII
Physics
Periodic Motion
Question
A particle executing SHM has a maximum speed of
30
c
m
s
−
1
and a maximum acceleration of
60
c
m
s
−
2
. The period of oscillation is:
π
π
2
s
2
π
15
s
2
π
A
π
2
s
B
π
C
2
π
D
2
π
15
s
Open in App
Solution
Verified by Toppr
Maximum velocity
V
m
a
x
=
A
w
=
30
c
m
s
−
1
Maximum acceleration
a
m
a
x
=
A
2
w
=
60
c
m
s
−
1
Where
A
=
A
m
p
l
i
t
u
d
e
w
=
a
n
g
u
l
a
r
f
r
e
q
u
e
n
c
y
∴
A
2
w
=
60
⟹
A
×
30
=
60
⟹
A
=
2
c
m
∴
w
=
30
2
s
−
1
=
15
s
e
c
−
1
∴
T
=
2
π
w
=
2
π
15
s
e
c
Was this answer helpful?
8
Similar Questions
Q1
A particle executing SHM has a maximum speed of
30
c
m
s
−
1
and a maximum acceleration of
60
c
m
s
−
2
. The period of oscillation is:
View Solution
Q2
A particle executing SHM has a maximum speed of
0.5
m
s
−
1
and maximum acceleration of
1
m
s
−
2
.
The angular frequency of oscillation is:
View Solution
Q3
A particle executing SHM has maximum acceleration of
2
π
m/s
2
and maximum velocity of
6
m/s
. Find its time period
View Solution
Q4
A particle executing SHM has maximum acceleration of
2
π
m/s
2
and maximum velocity of
6
m/s
. Find its time period of vibration
View Solution
Q5
The maximum speed of a particle executing
S
H
M
is
1
m
/
s
and maximum acceleration is
1.57
m
/
s
2
. Its time period is
View Solution