The variation of displacement with time of a particle executing free simple harmonic moton is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

Option 1 -

a

Option 2 -

b

Option 3 -

c

Option 4 -

d

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V
Vishal Baghel

T = 2 π l g

2 = 2 π 2 g

g = 2 π 2 m / s 2

V
Vishal Baghel

Velocity of block in equilibrium, in first case,

v = A ω = A . k M

Velocity of block in equilibrium, is second case,

v ' = A ' ω ' = A ' k M + m

From conservation of momentum,

Mv = (M + m) v’

M A k M = ( M + m ) A ' k M + m A ' = A M M + m

V
Vishal Baghel

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

V
Vishal Baghel

Kindly consider the following figure

V
Vishal Baghel

K = U

½ mω² (A² - x²) = ½ mω²x²

A² - x² = x²

A² = 2x²

x = ± A/√2

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