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Physics Oscillations

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3 weeks ago

Physics Oscillations Class 11th

A block ' $A$ ' of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude ' $A$ ' on a frictionless horizontal plane. If it collides elastically with another block ' $B$ ' of mass $2 m$ kept initially at rest, when ' $A$ ' is passing through its equilibrium point, the amplitude of oscillation for the block ' $A$ ' become $f$ A. The value of $f$ is:

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Raj Pandey

Contributor-Level 9

$v_{m a x} = A ω = A \sqrt[]{\frac{k}{m}}$

Momentum conservation $m v_{max} + 0 = m v_{A} + 2 m v_{B}$

$v_{A} + 2 v_{B} = v_{m a x}$

$v_{2} - v_{1} = u_{1} - u_{2} \Rightarrow v_{B} - v_{A} = v_{m a x}$

By (1) & (2) $v_{A} = - \frac{A ω}{3}$

$|v_{A}| = A_{new} ω$

$\frac{A ω}{3} = A_{new} ω ∴ A_{new} = \frac{A}{3}; f = \frac{1}{3}$

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3 weeks ago

Physics Oscillations Oscillations

The fundamental frequency of an organ pipe open at one end is 300 Hz. The frequency of 3rd overtone of n Find n.*this organ pipe is 100 Hz

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Vishal Baghel

Contributor-Level 10

f? = 300 Hz
3rd overtone = 7f? = 2100 Hz

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Physics Oscillations Class 11th

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20cm. The length of the open organ pipe is……….cm.

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Raj Pandey

Contributor-Level 9

$H = \frac{λ}{4}$

$\Rightarrow 20 = \frac{λ}{4}$

$λ = 80 c m$

Now first overtone for open organ pipe

$L_{2} = 4 \frac{λ}{4} = λ = 80 c m$

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3 weeks 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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3 weeks ago

Physics Oscillations Class 11th

The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

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Vishal Baghel

Contributor-Level 10

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

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3 weeks ago

Physics Oscillations Class 11th

For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be

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Vishal Baghel

Contributor-Level 10

y = A sin (2πt/T)
t? - t? = (T/2π) [sin? ¹ (x? /A) - sin? ¹ (x? /A)]

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3 weeks ago

Physics Oscillations Class 11th

The x-t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at t=2 is:

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Raj Pandey

Contributor-Level 9

From x-t graph,

$\begin{array}{r} A = 1, T = 8 \\ \Rightarrow ω = \frac{2 π}{T} \\ \Rightarrow ω = \frac{π}{4} \\ at t = 2, x = 1 \\ a = - ω^{2} x \\ \Rightarrow a = \frac{- π^{2}}{16} * 1 \\ \Rightarrow a = \frac{- π^{2}}{16} m / s^{2} \end{array}$

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Physics Oscillations Class 11th

A series LCR circuit with inductance $10 H$ , capacitance $10 μ F$ , resistance $50 Ω$ is connected to an ac source of voltage, $V = 200 s i n ? (100 t)$ volt. If the resonant frequency of the LCR circuit is $v_{0}$ and the frequency of the ac source is $v$ , then

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Raj Pandey

Contributor-Level 9

$ω = 100$

$v = \frac{ω}{2 π} = \frac{100}{2 π} = \frac{50}{π} H z$

Resonance frequency

$v_{0} = \frac{1}{2 π \sqrt[]{L C}} = \frac{1}{2 π} \sqrt[]{\frac{1}{10 * 10 * 10^{- 6}}}$

$= \frac{50}{π} H z$

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4 weeks ago

Physics Oscillations Class 11th

A body is executing simple harmonic motion with frequency ' $n$ ', the frequency of its potential energy is:

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alok kumar singh

Contributor-Level 10

Displacement equation of SHM of frequency ' $n$ '

$x = A s i n ? (ω t) = A s i n ? (2 π n t)$

Now,

Potential energy $U = \frac{1}{2} {k x}^{2} = \frac{1}{2} {K A}^{2} {s i n}^{2} ? 2 π n t$

$= \frac{1}{2} {k A}^{2} [\frac{1 - c o s ? 2 π 2 n t}{2}]$

So frequency of potential energy $= 2 n$

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4 weeks ago

Physics Oscillations Class 11th

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

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Raj Pandey

Contributor-Level 9

$\begin{array}{r} v \propto \sqrt[]{T} \\ \frac{v_{i}}{v_{f}} = \sqrt[]{\frac{1}{2}} = \frac{1}{\sqrt[]{2}} \end{array}$

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