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A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v . The particle is stopped, then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2y. After this change
(a) the frequency of oscillation will change by a factor of
(A) 4.                                                                     (B)
(C) 2.                                                                   (C)
(b) the maximum displacement of the particle will change by a factor of
(A) 4.                                                                 (B)
(B) 2.                                                                    (D)
c) the magnitude of the maximum acceleration of the particle will change by a factor of
(A) 4.                                                                   (B)
(B) 2.                                                                      (D)
(E) 1 (it remains unchanged).
2 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							(a) The correct option is (E).
In simple harmonic motion,
1) The time period (T) of vibrations varies inversely as the square root of the force constant (k) of the spring.
2) The time period (T) of vibrations varies directly as the square root of the mass (m) of body attached to the string.
Thus the time period of an object of mass m on a spring executes a simple harmonic motion is given by,
232-946_1.PNG
Therefore the maximum displacement of the particle will change by a factor of 2. Thus option (C) is correct.
232-2202_1.PNG
2 years ago
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