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Physics Oscillations

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2 months ago

Physics Oscillations Oscillations

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is 3E/4 then its displacement 'y' is given by:

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Vishal Baghel

Contributor-Level 10

E = ½mω²a² . (i)
When KE = 3E/4, PE = E - KE = E/4
E/4 = ½mω²y² . (ii)
Divide eq? (i) by eq? (ii) we get
4 = a²/y² => y = a/2

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2 months ago

Physics Oscillations Class 11th

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of the bob, when the length makes an angle of 60° to the vertical will be _______ m/s. (g = 10 m/s²)

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Raj Pandey

Contributor-Level 9

(1/2)mu² = (1/2)mv² + mgl (1-cosθ)
⇒ v² = u² - 2gl (1-cosθ)
⇒ v² = 3² - 2*10*0.5* (1 - 1/2)
⇒ v² = 9 - 5 = 4
v = 2m/s

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2 months ago

Physics Oscillations Class 11th

In the reported figure, two bodies A and B of masses 200g and 800g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be _______ rad/s when k = 20 N/m.

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Raj Pandey

Contributor-Level 9

ω = √k_eq/μ [μ = (m? )/ (m? +m? ) (Reduced mass)]
k_eq = (k * 4k)/ (k+4k) = 4k/5
ω = √ (4k/5) / (m? / (m? +m? )
= √ (4*20/5) / (0.2*0.8)/ (0.2+0.8)
= √ (16/0.16)
= 10 rad/s

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2 months ago

Physics Oscillations Class 11th

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

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alok kumar singh

Contributor-Level 10

v = ω√ (A² - x²)
K.E. = ½mv² = ½mω² (A² - x²)
Total Energy T.E = ½mω²A²
At x = A/2
K.E. = ½mω² (A² - (A/2)²) = ½mω²A² (3/4)
K.E./T.E. = (½mω²A² (3/4) / (½mω²A²) = 3/4

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3 months ago

Physics Oscillations Class 11th

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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alok kumar singh

Contributor-Level 10

Velocity of block in equilibrium, in first case,

$v = A ω = A . \sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v' = A' ω' = A' \sqrt[]{\frac{k}{M + m}}$

From conservation of momentum,

Mv = (M + m) v'

$\Rightarrow M A \sqrt[]{\frac{k}{M}} = (M + m) A' \sqrt[]{\frac{k}{M + m}} \Rightarrow A' = A \sqrt[]{\frac{M}{M + m}}$

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New answer posted

3 months ago

Physics Oscillations Oscillation

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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Vishal Baghel

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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3 months ago

Physics Oscillations Class 11th

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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Vishal Baghel

Contributor-Level 10

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

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

= 7

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New answer posted

3 months ago

Physics Oscillations Class 11th

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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Vishal Baghel

Contributor-Level 10

Time period of second pendulum is 2 seconds. &nbs

...more

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3 months ago

Physics Oscillations Class 11th

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

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Vishal Baghel

Contributor-Level 10

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

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New answer posted

3 months ago

Physics Oscillations Class 11th

If two similar spring each of spring constant K₁ are joined in series, the new spring constant and time period would be changed by a factor

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Payal Gupta

Contributor-Level 10

T $1 = 2 π \sqrt[]{\frac{m}{k}}$

$k_{e q} = \frac{k * k}{k + k} = \frac{k}{2}$

$T_{2} = 2 π \sqrt[]{\frac{m}{(k / 2)}} = 2 π \sqrt[]{\frac{2 m}{k}}$

$\Rightarrow \frac{T_{2}}{T_{1}} = \sqrt[]{2} a n d \frac{k_{2}}{k_{1}} = \frac{1}{2}$

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