Question

A mass attached to a spring oscillates with a period of 3.15 s (a) If the mass starts from rest at x = 0.0480 m and time t = 0, where is it at time t = 6.45 s? m(b) Is the mass moving in the positive or negative x direction at t - 6.45 s? Opositive x direction O negative x direction Explain.

Answer 1

Given

Spring, mass system with time period T = 3.15 s

we know that the time period of an oscillator is equal to the time taken to complete one oscillation that is in completing one oscillation the mass will cover 4 times the amplitude ,

for e.g , if the mass started from one of the extream ends(say right side ) moves towards mean position and to the other extream end , from there again moves towards mean position finally reaches the same point from where it started oscillations .

from the given data

at t = 0 s , the mass starts from rest and which is at x = 0.0480 m at t = 0 s so we assume that the spring mass system is on horizontal frictionless surface means it starting from zero velocity with a stretch of x=0.048 m

so the mass is moving towards mean position after t=0 s ,that is in the -ve X direction

a) and in time t= 6.45 s , it will complete 6.45/3.15 = 2.048 complete oscillations

means by completion of two oscillations it reaches the same position so the displacement is zero , now the position of the mass in 0.48 oscillations is

   for that we should calculate the velocity of the oscillator (mass) as

   v = (4*0.0480)/3.15 m/s = 0.06095 m/s

so mass can cover a distance of x = 0.06095*0.48 m = 0.029256 m

and the direction of the mass is towards -ve X direction

Answers
(a) At t= 6.45 s the mass is at x = 0.029256 m from the extream right end

(b) moving in the Negative X direction