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

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]

[IMAGE 108]

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

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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]

[IMAGE 109]
[IMAGE 110]
[IMAGE 111]
[IMAGE 112]

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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]

[IMAGE 113]

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

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(1)

[IMAGE 114]

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]

[IMAGE 115]

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(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]

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(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]

[IMAGE 116]

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(20)

[IMAGE 117]

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

$m = 490 gm = 0.49 kg$

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

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

$Number of oscillations in 14 π is$

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

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]

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(67)

[IMAGE 118]

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

[IMAGE 119]

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

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

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

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

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]

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(99)

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

$From conservation of energy:$

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

[IMAGE 120]----------

$Maximum tension occurs at mean position.$

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

$∴$ $T = m g + 2 m g (1 - \cos θ)$

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

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

$∴$ $x = 99$

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]

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(20)

$T = 3.14 = π$

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

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

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

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