Physics Oscillations

If length of stretched string is increased by 50% keeping mass per unit length same, then in order to keep its frequency of vibration constant, its tension must be increases by

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Contributor-Level 10

$f \propto \frac{\sqrt[]{T}}{l}$

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

$\Rightarrow T \propto l^{2}$ for same frequency

l' = 1.5 l

$\Rightarrow T' = 2.25 T$ 0020

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In the diagram shown below, the block has charge $Q$ and is attached at one end of the light spring. The surface is smooth and horizontal and spring is in its natural length. If the electric field $E$ is switched on, the maximum elongation in the spring will be

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Contributor-Level 10

From work energy theorem

$W_{s p} + W_{E} = ? K$

$- \frac{1}{2} k x_{0}^{2} + Q E . x_{0} = 0$

$? x_{0} = \frac{2 Q E}{k}$

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A particle executes S.H.M. with amplitude 'a' and time period 'T'. 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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Kindly go through the solution

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

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

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

->a = 7

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Given below are two statements:

Statement I : A second's pendulum has a time period of 1 second.

Statement II : It takes precisely one second to move between the two extreme positions.

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

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Kindly go through the solution

Time period of second pendulum is 2 seconds.

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A particle executes S.H.M., the graph of velocity as a function of displacement is:

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-> $v = ω \sqrt[]{A^{2} - x^{2}} \Rightarrow \frac{v^{2}}{A^{2} ω^{2}} + \frac{x^{2}}{A^{2}} = 1$ Path is ellipse.

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The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as $U = \frac{α}{r^{10}} - \frac{β}{r^{5}} - 3$

Where, α and β are positive constants. The equilibrium between two atoms will be ${(\frac{2 α}{β})}^{\frac{a}{b}},$ where α =…………….

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For equilibrium,

$\frac{d U}{d r} = 0 \Rightarrow \frac{- 10 α}{r^{11}} + \frac{5 β}{r^{6}} = 0 \Rightarrow r = {(\frac{2 α}{β})}^{\frac{1}{5}}$

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A student is performing the experiment of resonance column. The diameter of the column tube is 6cm. THE Frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336m/s. The zero of the meter scale coincides with tip end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:

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Physics Oscillations Oscillations

Contributor-Level 10

For first resonance,

$l + 0.3 d = \frac{v}{4 f}$

$\Rightarrow l + 0.3 * 6 = \frac{336 * 100}{4 * 504} \Rightarrow l = 14.8 c m$

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If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

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Physics Oscillations Oscillations

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

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In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. New amplitude of oscillation will be:

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