A particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are both doubled then(a) the period of oscillation will change by a factor of(A) 4.                                                                    (B) (C) 2.                                                                     (D) (b) the maximum speed of the particle will change by a factor of(A) 4.                                                              (B) (C) 2.                                                              (D) (c) the magnitude of the maximum acceleration of the particle will change by a factor of(A) 4.                                                                 (B) (C) 2.                                                                  (D) (E) 1 (it remains unchanged).

Navjyot Kalra
8 years ago
a) The correct option is (D). 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.
So the time period of an object of mass m on a spring executes a simple harmonic motion is given by,
So, T = 2π √m/k
Thus in simple harmonic motion, time period of the oscillation is independent of the amplitude.
If the mass m of the particle is doubled, then the time period (T ') of the simple harmonic motion would be,
T ' = 2π √2m/k
So, T '/ T = (2π √2m/k)/( 2π √m/k)
= √2
T ' = √2T
From the above observation we conclude that, the period of oscillation will change by a factor of √2. Thus option (D) is correct.
The correct option is (D).The maximum speed vmax is given by,
vmax = xm √k/m
Here k is the spring constant, m is the mass of the oscillator and xm is the amplitude of oscillation.
(amax )new = amax
From the above observation we conclude that, the magnitude of maximum acceleration of the particle will change by a factor of 1. Thus option (E) is correct.