Question

The period of oscillation of a spring-and-mass system is 0.60 s and the amplitude is 4.1 cm. What is the magnitude of the acceleration at the point of maximum extension of the spring?

I have too many unknowns to use the equation for frequency, period, acceleration, velocity. What am I missing?? Please help...

Answer 1

y = A sin (2 pi t/T)
dy/dt = A(2 pi/T) cos (2 pi t/T)
d^2y/dt^2 = -A (2 pi/T)^2 sin ( 2 pi t/T)

when extension is max, sin (2 pi/T) = + or - 1

|d^2/dt^2 max| = A (2 pi/T)^2

= .041 (2 pi/ .6)^2

Answer 2

Won't the amplitude and period be greater at a maximum?? Do you know why we can use the given values??? As far as I can tell, we assume that 0.6 and 4.1 are maximums...I am not following. This gives me the correct answer, so, I guess I'm not understanding the equation or the question.

Answer 3

The amplitude is A, the maximum of deflection
when you say
y = A sin (2 pi t/T)
you are saying that y starts at zero, goes up to A when t/T = 1/4 or 2pi t/T = pi/2 = 90 degrees etc
then the velocity is dy/dt and has a max of A (2pi/T)
and the acceleration (d^2y/dt^2) has a max of A (2 pi/T)^2

Answer 4

The argument of the trig function can be written several ways, for example
(2 pi t/T), the way I did it because you were given T
or
(2 pi f t) because T = 1/f
or
( w t) where w is the Greek letter omega and is the radial frerquency, 2 pi f, in radians/second

in any case, when t = T, time is one period, the argument is 2 pi , a full circle, which is what it is all about

Answer 5

The amplitude and period are constants for simple harmonic motion. The deflection, y, changes with time and has minimum of -A and maximum of +A
if sin function is used with no phase angle as I did it, y is 0 at start, +A at t = T/4, 0 at t=T/2, -A at t = 3 T/4
then finally back to 0 at t = T

Answer 6

Now get your text out and read the chapter on harmonic motion carefully or Google "simple harmonic motion" right now.

Answer 7

ok. thanks.