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:

Vishal Baghel

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’

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

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)

Vishal Baghel

Kindly consider the following figure

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

Vishal Baghel

K = U

½ mω² (A² - x²) = ½ mω²x²

A² - x² = x²

A² = 2x²

x = ± A/√2

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:

Vishal Baghel

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

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