Oscillations NEET Previous Year Questions PDF Download

Q 1 :
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $ω (t)$ and average amplitude $A (t)$ of the system change with time $t$ . Which one of the following options schematically depicts these changes correctly? [2025]

[IMAGE 124]
[IMAGE 125]
[IMAGE 126]
[IMAGE 127]

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4

(4)

As the box oscillates, sand leaks slowly out of the box which leads to decrease in effective mass of the bob. As frequency is inversely proportional to the square root of $m$ , hence average frequency $(ω)$ increases and amplitude will decrease due to dissipation of energy.

Q 2 :
Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_{1}$ and $k_{2}$ , respectively, oscillate vertically. If their maximum speeds are the same, the ratio $(\frac{A_{Q}}{A_{P}})$ of the amplitude $A_{Q}$ of mass $Q$ to the amplitude $A_{P}$ of mass $P$ is [2025]

$\sqrt{\frac{k_{2}}{k_{1}}}$
$\sqrt{\frac{k_{1}}{k_{2}}}$
$\frac{k_{2}}{k_{1}}$
$\frac{k_{1}}{k_{2}}$

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(2)

For a spring-mass system, the angular frequency is given by

$ω = \sqrt{\frac{k}{m}}$

For body $P : v_{1 (max)} = ω_{1} A_{P} = \sqrt{\frac{k_{1}}{m}} A_{P}$ ...(i)

For body $Q : v_{2 (max)} = ω_{2} A_{Q} = \sqrt{\frac{k_{2}}{m}} A_{Q}$ ...(ii)

Equate the equation (i) and (ii),

$\sqrt{\frac{k_{1}}{m}} A_{P} = \sqrt{\frac{k_{2}}{m}} A_{Q}$

or $\sqrt{k_{1}} A_{P} = \sqrt{k_{2}} A_{Q}$

Ratio of the amplitudes of Q and P:

$\frac{A_{Q}}{A_{P}} = \sqrt{\frac{k_{1}}{k_{2}}}$

Q 3 :
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is [2022]

11
9
10
8

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(1)

Let $L_{1} = 121$ $cm = \frac{121}{100} = 1.21$ $m$

$L_{2} = 100$ $cm = \frac{100}{100} = 1$ $m$

Let $T_{1} =$ longer pendulum ; $T_{2} =$ shorter pendulum

We know, $T = 2 π \sqrt{\frac{L}{g}} \Rightarrow T \propto \sqrt{L}$

$\frac{T_{1}}{T_{2}} \propto \sqrt{\frac{L_{1}}{L_{2}}}$

$\frac{T_{1}}{T_{2}} \propto \sqrt{\frac{1.21}{1}} = \frac{1.1}{1}$

$10 T_{1} = 11 T_{2}$

10 vibrations of longer pendulum = 11 vibrations of shorter pendulum.

Q 4 :
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is [2021]

0.628 s
0.0628 s
6.28 s
3.14 s

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(1)

Force constant of spring, $k = \frac{F}{x}$

Here, $F = 10$ $N, x = 5$ $cm$ ; $k = \frac{10}{0.05} = 200$ $N/m$

If the mass of 2 kg is suspended by the spring, then period of oscillation is

$T = 2 π \sqrt{\frac{m}{k}}; T = 2 π \sqrt{\frac{2}{200}} = 2 π \times 0.1$

$T = 0.628$ $s$

Q 5 :
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20$ ${ms}^{- 2}$ at a distance of 5 m from the mean position. The time period of oscillation is [2018]

$2 π$ $s$
$π$ $s$
$2$ $s$
$1$ $s$

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(2)

Magnitude of acceleration of a particle moving in a S.H.M. is, $| a |$ $= ω^{2} y, where y is amplitude .$

$\Rightarrow 20 = ω^{2} (5) \Rightarrow ω = 2 {rad s}^{- 1}$

$∴$ $Time period of oscillation, T = \frac{2 π}{ω} = \frac{2 π}{2} = π s$

Q 6 :
A spring of force constant $k$ is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is $k'$ . Then they are connected in parallel and the force constant is $k''$ . Then $k' : k''$ is [2017]

1 : 9
1 : 11
1 : 14
1 : 6

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(2)

Let us assume, the length of spring be $l$ .

When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths $\frac{l}{6}, \frac{2 l}{6}$ and $\frac{3 l}{6}$ with force constant,

$k_{1} = \frac{k l}{l_{1}} = \frac{k l}{l / 6} = 6 k, k_{2} = \frac{k l}{l_{2}} = \frac{k l}{2 l / 6} = 3 k$ and $k_{3} = \frac{k l}{l_{3}} = \frac{k l}{3 l / 6} = 2 k$

When connected in series, $\frac{1}{k'} = \frac{1}{6 k} + \frac{1}{3 k} + \frac{1}{2 k} = \frac{1 + 2 + 3}{6 k} = \frac{1}{k}$

$∴$ $k' = k$

When connected in parallel, $k'' = 6 k + 3 k + 2 k = 11 k$

$∴$ $\frac{k'}{k''} = \frac{k}{11 k} = \frac{1}{11}$

Q 7 :
A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass $m$ is increased by 1 kg, the time period of oscillations becomes 5 s. The value of $m$ in kg is [2016]

$\frac{3}{4}$
$\frac{4}{3}$
$\frac{16}{9}$
$\frac{9}{16}$

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(4)

Time period of spring-block system, $T = 2 π \sqrt{\frac{m}{k}}$

For a given spring, $T \propto \sqrt{m}$

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$

Here, $T_{1} = 3 s, m_{1} = m, T_{2} = 5 s, m_{2} = m + 1, m = ?$

$\frac{3}{5} = \sqrt{\frac{m}{m + 1}} or \frac{9}{25} = \frac{m}{m + 1}$

$25 m = 9 m + 9 \Rightarrow 16 m = 9;$ $∴$ $m = \frac{9}{16}$ $kg$

Topic Question Set

Q 3 :
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is [2022]

Q 4 :
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is [2021]

Q 5 :
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20$ ${ms}^{- 2}$ at a distance of 5 m from the mean position. The time period of oscillation is [2018]

Q 6 :
A spring of force constant $k$ is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is $k'$ . Then they are connected in parallel and the force constant is $k''$ . Then $k' : k''$ is [2017]

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Topic Question Set

Q 2 : Two identical point masses P and Q, suspended from two separate massless springs of spring constants k1 and k2, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (AQAP) of the amplitude AQ of mass Q to the amplitude AP of mass P is [2025]

Q 3 : Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is [2022]

Q 4 : A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is [2021]

Q 5 : A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20ms-2 at a distance of 5 m from the mean position. The time period of oscillation is [2018]

Q 6 : A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and the force constant is k''. Then k':k'' is [2017]

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Q 3 :
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is [2022]

Q 4 :
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is [2021]

Q 5 :
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20$ ${ms}^{- 2}$ at a distance of 5 m from the mean position. The time period of oscillation is [2018]

Q 6 :
A spring of force constant $k$ is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is $k'$ . Then they are connected in parallel and the force constant is $k''$ . Then $k' : k''$ is [2017]