What is the magnitude of acceleration at the point of maximum extension of a spring?

Question

The period of oscillation of a spring-and-mass system is 0.51s and the amplitude is 4.9cm. What is the magnitude of the acceleration at the point of maximum extension of the spring?

Answer 1

The mass distance from the fixed point is A*sin(wt) where A=4.9 cm and w=2*pi/T where period T=0.51 s. Max distance (= max extension of a spring) will be at wt=pi=3.14. Since the acceleration is the second derivative of the distance it is -Aw^2*sin(wt) which at maximum extension of a spring (wt=pi=3.14) is Aw^2 (amplitude) that is 4.9*(2*pi/0.51)^2=743.73 cm/sec^2.

Answer 2

the accleration will be maximum at the maximum diestance from mean position.here w=2*pi/T
where pi=22/7
T=time period=0.51sec
a=amplitude=4.9cm
equation of the motion is given by
x=asin(wt)
acceleration = -w^2x
magnitude of accelration:ie maximum acceleration= -w^2

Answer 3

Zero.

At the point of maximum extension, the spring is in the process of reversing its direction - it has stopped.

Answer 4

x = 4.9cos((2π/0.51)t)
v = -(9.8π/0.51)sin((2π/0.51)t)
a = -(19.6π²/0.2601)cos((2π/0.51)t)
maximum elongation @ t = 0
a(0) = -743.73 cm/sec²

Answer 5

F = Cu and F = ma, so a = Cu/m
a(max) = {Cu(max)}/m and u(max) = A
C * T^2 =(2pi)^2 * m
Calculate yourself

Th