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Oscillations

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a month ago

Oscillations Class 11th

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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Vishal Baghel

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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New answer posted

a month ago

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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Vishal Baghel

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

a month ago

Oscillations Class 11th

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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New answer posted

a month ago

Oscillations Class 11th

Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T_b / T_a = √x, where value of x is . (Round off to the Nearest Integer)

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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New question posted

a month ago

Oscillations Class 11th

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New answer posted

a month ago

Oscillations Class 11th

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

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Vishal Baghel

Contributor-Level 10

K = U

½ mω² (A² - x²) = ½ mω²x²

A² - x² = x²

A² = 2x²

x = ± A/√2

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New answer posted

2 months ago

Oscillations Class 11th

An object of mass m is suspended at the end of a massless wire of length L and area of cross-selection, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:

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Vishal Baghel

Contributor-Level 10

f = (1/2π)√ (k/m), k=YA/L. f= (1/2π)√ (YA/mL)

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New answer posted

2 months ago

Oscillations Class 11th

An object of mass 0.5 kg executing simple harmonic motion. Its amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = T/4 s starting from mean position. Assume that the initial phase of the oscillation is zero.

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Vishal Baghel

Contributor-Level 10

T = 2π√ (m/k)
0.2 = 2π√ (0.5/k)
k = (0.5) (2π/0.2)² = 50π² ≈ 500
x = A sin (ωt) = 5 sin (2π/T * t)
At t=T/4, x = 5 sin (π/2) = 5cm
PE = ½kx² = ½ * 500 * (0.05)² = 250 * 0.0025 = 0.625 J

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New answer posted

2 months ago

Oscillations Class 11th

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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New answer posted

2 months ago

Oscillations Class 11th

When a particle executes SHM, the nature of graphical representation of velocity as a function of

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Vishal Baghel

Contributor-Level 10

Let x = A sin $ω t$ & v = $A ω c o s ω t$

$∴ v = ω \sqrt[]{A^{2} - x^{2}}$

$v^{2} = ω^{2} x^{2} - ω^{2} A^{2}$

$\frac{v^{2}}{ω^{2} A^{2}} + \frac{x^{2}}{A^{2}} = 1$

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