33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Oscillations
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113
When connected in series,
1 1 1 1 1 2 3 1
k′ = 6k + 3k + 2k = + + =
6k k
\ k′ = k
When connected in parallel,
k′′ = 6k + 3k + 2k = 11k
k′
k
\
k′′ = 11k
= 1
11
42. (d) : Time period of spring - block system,
m
T = 2π
k
For given spring, T ∝
T1
m1
=
T m
2
2
m
Here, T 1 = 3 s, m 1 = m, T 2 = 5 s, m 2 = m + 1, m = ?
3 m 9 m
= or =
5 m + 1 25 m + 1
25m = 9m + 9 ⇒ 16m = 9; ∴ m =
9
16 kg
43. (d) : A mass M is suspended from a massless spring
of spring constant k as shown in figure (a). Then,
time period of oscillation is
M
T = 2π ...(i)
k
When a another mass M is also suspended with it
as shown in figure (b). Then, time period of
oscillation is,
M + M
T ′ = 2π
= 2π
k
2M
⎛ ⎞
= 2
⎝
⎜2π
M ⎠
⎟ = 2T ( Using(i) )
k
44. (c) : For simple harmonic motion,
v= ω a 2 −x
2 .
2
a a
When x = v = a
2 3 2
, ω − = ω a .
2 4 4
π π π
As ω = 2 ∴ = 2 ⋅ 3
3a
, v a ⇒ v=
.
T T 2
T
45. (a) :
k
If the restoring force mAw 2 > mg, then the mass will move
up with acceleration, detached from the pan.
g
20
ie .., A > ⇒ A > > 010 . m.
k/
m
200
The amplitude > 10 cm.
i.e. the minimum is just greater than 10 cm.
(The actual compression will include x 0 also. But when
talking of amplitude, it is always from the equilibrium
position with respect to which the mass is oscillating.
46. (c) : Let l be the length of block immersed in liquid
as shown in the figure.
When the block is floating,
\ mg = Alrg
If the block is given vertical displacement y then the
effective restoring force is
F = –[A(l + y)rg – mg] = –[A(l + y)rg – Alrg]
= –Argy
Restoring force = –[Arg]y. As this F is directed towards
equilibrium position of block, so it will execute simple
harmonic motion.
Here inertia factor = mass of block = m
Spring factor = Arg
m
∴ Time period, T = ∝
A g ie T 1
2π
.., .
ρ
A
47. (d) : When the spring joined in series, the total
extension in spring is
⇒ = + = − F F
y y1 y2
−
k1 k
⇒ =− ⎡ 1 + 1 ⎤
y F⎢
⎥
2 ⎣ k 1 k 2⎦
Thus spring constant in this case becomes
kk 1 2
k =
k1+
k2
48. (c) : Let k be the force constant of spring. If k′ is the
force constant of each part, then
1 4
= ⇒ k′ = 4k.
k k ′
m 1 m T
∴ Time period = 2π
= × 2π
= .
4k
2 k 2
49. (b) : The time period of a spring mass
system as shown in figure 1 is given by
T = 2π m/ k, where k is the spring constant.
∴ t = 2π m/ k ...(i)
1 1
and t2 = 2π m/ k ...(ii)
2
The spring has a length l. Whem m is placed over it, the
equilibrium position becomes O′.
If it is pressed from O′ (the equilibrium position) to O′′,
O′O′′ is the amplitude.
\ OO ′ = mg
k
= 2 × 10
200
=
mg = kx 0 .
010 . m.
Now, when they are connected in parallel as shown in
figure 2(a), the system can be replaced by a single spring
of spring constant, k eff = k 1 + k 2 , as shown in figure 2(b).
Since mg = k 1 x + k 2 x = k eff x