Oscillation and Simple Harmonic Motion | Physics JEE previous year questions with Solutions

Q 1 :

A simple pendulum doing small oscillations at a place R height above earth surface has time period of $T_{1} = 4 s .$ $T_{2}$ would be it's time period if it is brought to a point which is at a height 2R from earth surface.

Choose the correct relation [R = radius of earth] [2024]

$2 T_{1} = 3 T_{2}$
$2 T_{1} = T_{2}$
$T_{1} = T_{2}$
$3 T_{1} = 2 T_{2}$

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(4)

$T = 2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{l r^{2}}{GM}} = 2 π \sqrt{\frac{l}{GM}} r$

$\Rightarrow \frac{T_{2}}{T_{1}} = \frac{r_{2}}{r_{1}} = \frac{3 R}{2 R} \Rightarrow 2 T_{2} = 3 T_{1}$

Q 2 :

A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency $f_{1}$ . The frequency of oscillations if a mass 9 m is suspended from the same spring is $f_{2} .$ The value of $\frac{f_{1}}{f_{2}}$ is _______ . [2024]

(3)

Q 3 :

In simple harmonic motion, the total mechanical energy of a given system is E. If the mass of oscillating particle P is doubled, then the new energy of the system for same amplitude is [2024]

$E \sqrt{2}$
$E$
$2 E$
$E / \sqrt{2}$

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(2)

Total energy $E = K E + P E$

$E = \frac{1}{2} m ω^{2} (A^{2} - x^{2}) + \frac{1}{2} m ω^{2} x^{2}$

$E = \frac{1}{2} k A^{2}, since A is same, T . E . will be same$

Q 4 :

The time period of simple harmonic motion of mass M in the given figure is $π \sqrt{\frac{α M}{5 k}},$ where the value of $α$ is ________ . [2024]

(12)

$k_{eq} = \frac{2 k \cdot k}{3 k} + k = \frac{5 k}{3}$

Angular frequency of oscillation $(ω) = \sqrt{\frac{k_{eq}}{m}}$

$(ω) = \sqrt{\frac{5 k}{3 m}}$

Period of oscillation $(τ) = \frac{2 π}{ω} = 2 π \sqrt{\frac{3 m}{5 k}}$

$= π \sqrt{\frac{12 m}{5 k}}$

Q 5 :

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.

Reasosn (R) : The mass of the pendulum remains unchanged at Earth and the other planet.

In the light of the above statements, choose the correct answer from the options given below: [2025]

Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(A) is true but (R) is false.
(A) is false but (R) is true.
Both (A) and (R) are true and (R) is the correct explanation of (A).

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(1)

$g = \frac{G M}{R^{2}}$

$g' = \frac{G (4 M)}{{(2 R)}^{2}} = g$

A is correct, R is correct; but since $T = 2 π \sqrt{\frac{l}{g}}$

Doesn't depend on mass; R doesn't explain A.

Q 6 :

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.

Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.

In the light of the above statements, choose the most appropriate answer from the options given below: [2025]

Both (A) and (R) are true but (R) is not the correct explanation of (A).
Both (A) and (R) are true and (R) is the correct explanation of (A).
(A) is true but (R) is false.
(A) is false but (R) is true.

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(2)

As h increases, g decreases, T increases

$T = 2 π \sqrt{\frac{l}{g}}$

$g = \frac{g_{0} R^{2}}{{(R + h)}^{2}}$

Q 7 :

Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_{1}$ and $k_{2}$ , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is [2025]

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

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(1)

Here $ω = \sqrt{\frac{k}{m}}$ and maximum velocity $V = A ω = A \sqrt{\frac{k}{m}}$

So, $\frac{V_{A}}{V_{B}} = \sqrt{\frac{k_{1}}{k_{2}}}$

Q 8 :

Two blocks of masses m and M, (M > m), are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released then ( $μ$ = coefficient of friction between the two blocks)

(A) The time period of small oscillation of the two blocks is $T = 2 π \sqrt{\frac{(m + M)}{k}}$

(B) The acceleration of the blocks is $a = \frac{k x}{M + m}$

(x = displacement of the blocks from the mean position)

(C) The magnitude of the frictional force on the upper block is $\frac{m μ | x |}{M + m}$

(D) The maximum amplitude of the upper block, if it does not slip, is $\frac{μ (M + m) g}{k}$

(E) Maximum frictional force can be $μ (M + m) g$

Choose the correct answer from the options given below: [2025]

A, B, D Only
B, C, D Only
C, D, E Only
A, B, C Only

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(1)

A. Assuming no slipping, $T = 2 π \sqrt{\frac{m_{total}}{k}} = 2 π \sqrt{\frac{m + M}{k}} A$ is correct.

B. Let block is displaced by x in (+ve) direction so force on block will be in (–ve) direction

$F = - K x \Rightarrow (M + m) a = - K x$

$\Rightarrow a = - \frac{K x}{(M + m)}$ , B is correct.

C. As upper block is moving due to friction thus

$f = m a = \frac{m K x}{(M + m)}$ , C is not correct.

D. If both the blocks moves together at maximum amplitude, the friction force on the block of mass m is also maximum,

For no slipping $\frac{m k A}{m + M} \leq μ m g$

$\Rightarrow A = \frac{μ (M + m) g}{K}$ , D is correct.

E. Maximum friction, $f_{\max} = μ m g$ , E is incorrect.

Q 9 :

Two simple pendulums having lengths $l_{1}$ and $l_{2}$ with negligible string mass undergo angular displacements $θ_{1}$ and $θ_{2}$ , from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct? [2025]

$θ_{1} l_{2}^{2} = θ_{2} l_{1}^{2}$
$θ_{1} l_{1} = θ_{2} l_{2}$
$θ_{1} l_{1}^{2} = θ_{2} l_{2}^{2}$
$θ_{1} l_{2} = θ_{2} l_{1}$

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(4)

Angular frequency of simple pendulum, $ω = \sqrt{\frac{g}{l}}$

Angular acceleration, $α = - ω^{2} θ$

$∴ \frac{g}{l_{1}} θ_{1} = \frac{g}{l_{2}} θ_{2}$

$\Rightarrow θ_{1} l_{2} = θ_{2} l_{1}$

Q 10 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. [2023]

Assertion A: A pendulum clock when taken to Mount Everest becomes fast.

Reason R: The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.

In the light of the above statements, choose the most appropriate answer from the options given below:

Both A and R are correct but R is NOT the correct explanation of A
Both A and R are correct and R is the correct explanation of A
A is not correct but R is correct
A is correct but R is not correct

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(3)

When we go on the Mount Everest the value of gravitational acceleration decreases $(g = \frac{g_{0}}{{(1 + \frac{h}{R_{e}})}^{2}})$

Therefore, the time period of oscillation $(T = 2 π \sqrt{\frac{ℓ}{g}})$ increases and the pendulum clock becomes slow, thus the assertion is wrong but the reason is correct.

Q 11 :

$' T'$ is the time period of a simple pendulum on the earth’s surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth’s radius) above the earth’s surface. Then, the value of $' x'$ will be [2023]

$\frac{1}{4}$
$4$
$\frac{1}{2}$
$2$

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(4)

$T = 2 π \sqrt{\frac{l}{g}}$

$g_{h} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + R})}^{2} = \frac{g}{4}$

$T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = T \sqrt{\frac{g}{g_{h}}} = 2 T$

Q 12 :

For a simple harmonic motion in a mass–spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is $ω_{1}$ . When the mass of the block is 2 kg, the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is [2023]

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

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(2)

$ω = \sqrt{\frac{k}{m}}$

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

Q 13 :

Choose the correct length $(L)$ versus square of time period $(T^{2})$ graph for a simple pendulum executing simple harmonic motion. [2023]

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(3)

$T = 2 π \sqrt{\frac{l}{g}}$

$\Rightarrow T^{2} = \frac{4 π^{2}}{g} \times l$

$\Rightarrow T^{2} \propto l$

Q 14 :

A mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_{1}$ and $K_{2}$ . For the frictionless surface, the time period of oscillation of mass $m$ is [2023]

$2 π \sqrt{\frac{m}{K_{1} + K_{2}}}$
$\frac{1}{2 π} \sqrt{\frac{K_{1} - K_{2}}{m}}$
$\frac{1}{2 π} \sqrt{\frac{K_{1} + K_{2}}{m}}$
$2 π \sqrt{\frac{m}{K_{1} - K_{2}}}$

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(1)

On displacing mass $m$ to the right by $x$ ,

$F = - (k_{1} x + k_{2} x) = - (k_{1} + k_{2}) x$

$a = \frac{F}{m} = - (\frac{k_{1} + k_{2}}{m}) x = - ω^{2} x$

$∴$ $ω = \sqrt{\frac{k_{1} + k_{2}}{m}} \Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{k_{1} + k_{2}}}$

Q 15 :

A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{π}{\sqrt{x}}$ in SI unit. The value of $x$ is ________. [2023]

(5)

$F = - 2 k x, a = - \frac{2 k x}{m},$

$ω = \sqrt{\frac{2 k}{m}} = \sqrt{\frac{2 \times 20}{2}} = \sqrt{20} rad/s$

$T = \frac{2 π}{ω} = \frac{2 π}{\sqrt{10}} = \frac{π}{\sqrt{5}} \Rightarrow x = 5$

Q 16 :

A mass $m$ attached to the free end of a spring executes SHM with a period of 1 s. If the mass is increased by 3 kg, the period of oscillation increases by one second, the value of mass $m$ is ______ kg. [2023]

(1)

$T = 2 π \sqrt{\frac{m}{k}} = 1$

$T' = 2 π \sqrt{\frac{m + 3}{k}} = 2$

$\frac{T}{T'} = \sqrt{\frac{m}{m + 3}} = \frac{1}{2}$

$\Rightarrow \frac{m}{m + 3} = \frac{1}{4} \Rightarrow m = 1$

Q 17 :

In the figure given below, a block of mass $M = 490 g$ placed on a frictionless table is connected with two springs having the same spring constant $(K = 2 {Nm}^{- 1}) .$ If the block is horizontally displaced through $' X' m,$ then the number of complete oscillations it will make in $14 π$ seconds will be ______ . [2023]

(20)

$K_{eff} = K + K (as both springs are in use in parallel) = 2 k = 2 \times 2 = 4 N/m$

$m = 490 gm = 0.49 kg$

$T = 2 π \sqrt{\frac{m}{K_{eff}}} = 2 π \sqrt{\frac{0.49}{4}}$

$= 2 π \sqrt{\frac{49}{400}} = 2 π \frac{7}{20} = \frac{7 π}{10}$

$Number of oscillations in 14 π is$

$N = \frac{time}{T} = \frac{14 π}{7 π / 10} = 20$

Q 18 :

A block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 cm$ from its equilibrium position (at $x = 0$ ) on a frictionless surface from rest. The energy of the block at $x = 5 cm$ is $0.25 J.$ The spring constant of the spring is ________ ${Nm}^{- 1}$ . [2023]

(67)

$U_{i} = \frac{1}{2} k x_{0}^{2} and K_{i} = 0$

$U_{f} = \frac{1}{2} k {(\frac{x_{0}}{2})}^{2} and K_{f} = 0.25 J$

$\frac{1}{2} k x_{0}^{2} + 0 = \frac{1}{2} k \frac{x_{0}^{2}}{4} + 0.25$

$\frac{1}{2} k x_{0}^{2} \cdot \frac{3}{4} = \frac{1}{4}$

$\frac{1}{2} k \cdot \frac{3}{100} = 1 \Rightarrow k = \frac{200}{3} N/m \approx 67 N/m$

Q 19 :

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be $\frac{x}{40} N .$ The value of $x$ is ________ . [2023]

(99)

$\sin θ_{c} = \frac{A}{l} = \frac{10}{100} = \frac{1}{10}$

$From conservation of energy:$

$\frac{1}{2} m v^{2} = m g l (1 - \cos θ)$

$Maximum tension occurs at mean position.$

$∴$ $T - m g = \frac{m v^{2}}{l} \Rightarrow T = m g + \frac{m v^{2}}{l}$

$∴$ $T = m g + 2 m g (1 - \cos θ)$

$= m g [1 + 2 (1 - \sqrt{1 - \sin^{2} θ})] = m g [3 - 2 \sqrt{1 - \frac{1}{100}}]$

$= \frac{250}{1000} \times 9.8 [3 - 2 (1 - \frac{1}{200})] = \frac{99}{40}$

$∴$ $x = 99$

Q 20 :

A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s. The maximum force exerted by the spring on the block is ______ N. [2023]

(20)

$T = 3.14 = π$

$T = π = \frac{2 π}{ω} \Rightarrow ω = 2$

$F_{\max} = m a_{\max} = m (A ω^{2}) = m A {(2)}^{2}$

$= 5 \times 1 \times 4 = 20 N$

Q 21 :

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ______ cm. [Assume that the mass of the pendulum remains same.] [2026]

7.5
0.75
15
120

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(1)

$Time period becomes half$

and $T \propto \sqrt{ℓ}$

$So, length ℓ becomes \frac{ℓ}{4}$

$So, \frac{ℓ}{4} = \frac{30}{4} = 7.5$

Q 22 :

Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in the figure. The energy stored in A is E. The energy stored in B, when spring constants $k_{A}$ and $k_{B}$ of A and B, respectively satisfy the relation $4 k_{A} = 3 k_{B}$ is: [2026]

$2 E$
$4 E$
$\frac{4}{3} E$
$3 E$

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(4)

$For equilibrium k x = m g$

$U = \frac{1}{2} k x^{2} = \frac{1}{2} \frac{m^{2} g^{2}}{k}$

$U \propto \frac{m^{2}}{k}$

$\frac{U_{A}}{U_{B}} = {(\frac{m_{A}}{m_{B}})}^{2} (\frac{k_{B}}{k_{A}}) = {(\frac{1}{2})}^{2} (\frac{4}{3}) = \frac{1}{3}$

$\frac{E}{U_{B}} = \frac{1}{3} \Rightarrow U_{B} = 3 E$

Q 23 :

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their lengths is 1 : 3, then the force constant of the smaller piece is ______ N/m. [2026]

20
60
45
15

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(2)

$K ℓ = constant$

$K ℓ = K' (\frac{ℓ}{4})$

$K' = 4 K$

$K' = 60 N/m$

Q 24 :

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is: [2026]

30
20
27
5

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(1)

$μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{1 \times 0.2}{1.2}$

$μ = \frac{1}{6}$

$ω = \sqrt{\frac{k}{μ}} = \sqrt{\frac{150}{1 / 6}} = 30$

Topic Question Set

Q 1 :

A simple pendulum doing small oscillations at a place R height above earth surface has time period of $T_{1} = 4 s .$ $T_{2}$ would be it's time period if it is brought to a point which is at a height 2R from earth surface.

Choose the correct relation [R = radius of earth] [2024]

Q 2 :

A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency $f_{1}$ . The frequency of oscillations if a mass 9 m is suspended from the same spring is $f_{2} .$ The value of $\frac{f_{1}}{f_{2}}$ is _______ . [2024]

Q 3 :

In simple harmonic motion, the total mechanical energy of a given system is E. If the mass of oscillating particle P is doubled, then the new energy of the system for same amplitude is [2024]

Q 4 :

The time period of simple harmonic motion of mass M in the given figure is $π \sqrt{\frac{α M}{5 k}},$ where the value of $α$ is ________ . [2024]

Q 7 :

Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_{1}$ and $k_{2}$ , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is [2025]

Q 9 :

Two simple pendulums having lengths $l_{1}$ and $l_{2}$ with negligible string mass undergo angular displacements $θ_{1}$ and $θ_{2}$ , from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct? [2025]

Q 11 :

$' T'$ is the time period of a simple pendulum on the earth’s surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth’s radius) above the earth’s surface. Then, the value of $' x'$ will be [2023]

Q 12 :

For a simple harmonic motion in a mass–spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is $ω_{1}$ . When the mass of the block is 2 kg, the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is [2023]

Q 13 :

Choose the correct length $(L)$ versus square of time period $(T^{2})$ graph for a simple pendulum executing simple harmonic motion. [2023]

Q 14 :

A mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_{1}$ and $K_{2}$ . For the frictionless surface, the time period of oscillation of mass $m$ is [2023]

Q 16 :

A mass $m$ attached to the free end of a spring executes SHM with a period of 1 s. If the mass is increased by 3 kg, the period of oscillation increases by one second, the value of mass $m$ is ______ kg. [2023]

Q 18 :

A block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 cm$ from its equilibrium position (at $x = 0$ ) on a frictionless surface from rest. The energy of the block at $x = 5 cm$ is $0.25 J.$ The spring constant of the spring is ________ ${Nm}^{- 1}$ . [2023]

Q 19 :

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be $\frac{x}{40} N .$ The value of $x$ is ________ . [2023]

Q 20 :

A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s. The maximum force exerted by the spring on the block is ______ N. [2023]

Q 21 :

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ______ cm. [Assume that the mass of the pendulum remains same.] [2026]

Q 22 :

Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in the figure. The energy stored in A is E. The energy stored in B, when spring constants $k_{A}$ and $k_{B}$ of A and B, respectively satisfy the relation $4 k_{A} = 3 k_{B}$ is: [2026]

Q 23 :

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their lengths is 1 : 3, then the force constant of the smaller piece is ______ N/m. [2026]

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Topic Question Set

Q 1 : A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1=4s. T2 would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth] [2024]

Q 2 : A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency f1. The frequency of oscillations if a mass 9 m is suspended from the same spring is f2. The value of f1f2 is _______ . [2024]

Q 3 : In simple harmonic motion, the total mechanical energy of a given system is E. If the mass of oscillating particle P is doubled, then the new energy of the system for same amplitude is [2024]

Q 4 : The time period of simple harmonic motion of mass M in the given figure is παM5k, where the value of α is ________ . [2024]

Q 7 : Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is [2025]

Q 9 : Two simple pendulums having lengths l1 and l2 with negligible string mass undergo angular displacements θ1 and θ2, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct? [2025]

Q 11 : 'T' is the time period of a simple pendulum on the earth’s surface. Its time period becomes xT when taken to a height R (equal to earth’s radius) above the earth’s surface. Then, the value of 'x' will be [2023]

Q 12 : For a simple harmonic motion in a mass–spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is ω1. When the mass of the block is 2 kg, the angular frequency is ω2. The ratio ω2/ω1 is [2023]

Q 13 : Choose the correct length (L) versus square of time period (T2) graph for a simple pendulum executing simple harmonic motion. [2023]

Q 14 : A mass m is attached to two springs as shown in the figure. The spring constants of the two springs are K1 and K2. For the frictionless surface, the time period of oscillation of mass m is [2023]

Q 16 : A mass m attached to the free end of a spring executes SHM with a period of 1 s. If the mass is increased by 3 kg, the period of oscillation increases by one second, the value of mass m is ______ kg. [2023]

Q 18 : A block is fastened to a horizontal spring. The block is pulled to a distance x=10 cm from its equilibrium position (at x=0) on a frictionless surface from rest. The energy of the block at x=5 cm is 0.25 J. The spring constant of the spring is ________ Nm-1. [2023]

Q 19 : A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be x40 N. The value of x is ________ . [2023]

Q 20 : A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s. The maximum force exerted by the spring on the block is ______ N. [2023]

Q 21 : A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ______ cm. [Assume that the mass of the pendulum remains same.] [2026]

Q 22 : Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in the figure. The energy stored in A is E. The energy stored in B, when spring constants kA and kB of A and B, respectively satisfy the relation 4kA=3kB is: [2026]

Q 23 : A spring of force constant 15 N/m is cut into two pieces. If the ratio of their lengths is 1 : 3, then the force constant of the smaller piece is ______ N/m. [2026]

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Q 1 :

A simple pendulum doing small oscillations at a place R height above earth surface has time period of $T_{1} = 4 s .$ $T_{2}$ would be it's time period if it is brought to a point which is at a height 2R from earth surface.

Choose the correct relation [R = radius of earth] [2024]

Q 2 :

A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency $f_{1}$ . The frequency of oscillations if a mass 9 m is suspended from the same spring is $f_{2} .$ The value of $\frac{f_{1}}{f_{2}}$ is _______ . [2024]

Q 3 :

In simple harmonic motion, the total mechanical energy of a given system is E. If the mass of oscillating particle P is doubled, then the new energy of the system for same amplitude is [2024]

Q 4 :

The time period of simple harmonic motion of mass M in the given figure is $π \sqrt{\frac{α M}{5 k}},$ where the value of $α$ is ________ . [2024]

Q 7 :

Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_{1}$ and $k_{2}$ , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is [2025]

Q 9 :

Two simple pendulums having lengths $l_{1}$ and $l_{2}$ with negligible string mass undergo angular displacements $θ_{1}$ and $θ_{2}$ , from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct? [2025]

Q 11 :

$' T'$ is the time period of a simple pendulum on the earth’s surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth’s radius) above the earth’s surface. Then, the value of $' x'$ will be [2023]

Q 12 :

For a simple harmonic motion in a mass–spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is $ω_{1}$ . When the mass of the block is 2 kg, the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is [2023]

Q 13 :

Choose the correct length $(L)$ versus square of time period $(T^{2})$ graph for a simple pendulum executing simple harmonic motion. [2023]

Q 14 :

A mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_{1}$ and $K_{2}$ . For the frictionless surface, the time period of oscillation of mass $m$ is [2023]

Q 16 :

A mass $m$ attached to the free end of a spring executes SHM with a period of 1 s. If the mass is increased by 3 kg, the period of oscillation increases by one second, the value of mass $m$ is ______ kg. [2023]

Q 18 :

A block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 cm$ from its equilibrium position (at $x = 0$ ) on a frictionless surface from rest. The energy of the block at $x = 5 cm$ is $0.25 J.$ The spring constant of the spring is ________ ${Nm}^{- 1}$ . [2023]

Q 19 :

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be $\frac{x}{40} N .$ The value of $x$ is ________ . [2023]

Q 20 :

A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s. The maximum force exerted by the spring on the block is ______ N. [2023]

Q 21 :

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ______ cm. [Assume that the mass of the pendulum remains same.] [2026]

Q 22 :

Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in the figure. The energy stored in A is E. The energy stored in B, when spring constants $k_{A}$ and $k_{B}$ of A and B, respectively satisfy the relation $4 k_{A} = 3 k_{B}$ is: [2026]

Q 23 :

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their lengths is 1 : 3, then the force constant of the smaller piece is ______ N/m. [2026]