Oscillations

In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:

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Contributor-Level 10

Here, $K_{e q} = 2 k$ $_{}$

$T = 2 π \sqrt[]{\frac{m}{2 k}}$

$∴ f = \frac{1}{2 π} \sqrt[]{\frac{2 k}{m}}$

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a year ago

A bob of mass 'm' suspended by a thread of length I undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by 1/3^rd of the original length, then the time period of the simple harmonic oscillations will be:-

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Contributor-Level 10

Time period T = 2π $\sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T_{i} = T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow T_{f} = T' = 2 π \sqrt[]{\frac{4 l / 3}{g (1 - \frac{ρ}{σ})}} = 2 π \sqrt[]{\frac{16 l}{9 g}}$

$\Rightarrow T' = \frac{4}{3} T$

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a year ago

The variation of displacement with time of a particle executing free simple harmonic moton is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

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Potential Energy is maximum at extreme position

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a year ago

The displacement simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is:

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$x = A s i n ω t$

Angle travelled by while moving from A to B

$θ = 30 ° = \frac{π °}{6}$

$θ = ω t$

$\frac{π}{6} = ω * 3 [t = 3 s e c]$

$ω = \frac{π}{18}$

$\Rightarrow T = \frac{2 π}{ω} = \frac{2 π * 18}{π} = 36 s e c$

T = 36 sec

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a year ago

In figure (A), mass '2 m' is fixed on mass 'm' which is attached to two spring of spring constant k. In figure (B), mass 'm' is attached to two springs of spring constant 'k' and '2k'. If mass 'm' in (A) and in (B) are displaced by distance 'x' horizontally and then released, then time period T₁ and T₂ corresponding to (A) and (B) respectively follow the relation.

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Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Contributor-Level 10

$T_{1} = 2 π \sqrt[]{\frac{3 m}{2 k}}$

$T_{2} = 2 π \sqrt[]{\frac{m}{3 k}}$

$\frac{T_{1}}{T_{2}} = \frac{2 π \sqrt[]{\frac{3 m}{2 k}}}{2 π \sqrt[]{\frac{m}{3 k}}} = \frac{3}{\sqrt[]{2}}$

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a year ago

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________* 10^-2mm.

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Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Contributor-Level 10

Let 'x' he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

10 y = 9 x

$y = \frac{9 x}{10}$

Least count = $x - \frac{9 x}{10} = \frac{x}{10} = \frac{0.05}{10}$

= 0.005 cm

= 5 * 10^-2 mm

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a year ago

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is

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Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Contributor-Level 10

Volume = constant

$A l = C$

$A d l + l d A = 0$

$\frac{d l}{l} + \frac{d A}{A} = 0$

$\frac{d R}{R} * 100 = (\frac{d l}{l} - \frac{d A}{A}) * 100$

= 0.4 (0.4)

= 0.8 %

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a year ago

A particle executes S.H.M. with amplitude â€˜aâ€™ and time period â€˜Tâ€™. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt[]{x} a}{2} .$ The value of x is

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Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Contributor-Level 10

$v = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow \frac{A ω}{2} = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow s = \frac{\sqrt[]{3} A}{2} \Rightarrow x = 3$

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a year ago

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{α}{β} T .$ The value of a is

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Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Contributor-Level 10

Total oscillation = $\frac{1}{2} + \frac{1}{8} = \frac{5}{8}$

$\Rightarrow t i m e t a k e n = \frac{T}{2} + \frac{T}{12} = \frac{7 T}{12}$

= 7

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a year ago

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be:

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