A hanging spring stretches by 39.0 cm when an object of mass 490 g is hung on it

Question

A hanging spring stretches by 39.0 cm when an object of mass 490 g is hung on it from rest. The object is then pulled down an additional 15.5 cm and released from rest to oscillate without friction.

a.) determine the position x at a moment 84.4 s later.

b.) determine the distance traveled by the vibrating object

c.) another hanging spring stretches by 39.5 cm when an object of mass 480 g is hung on it from rest. The object is also pulled down an additional 15.5 cm and released from rest to oscillate without friction. Determine its position x at a time 84.4 s later.

d.) determine the distance traveled by the vibrating object.

Explanation / Answer

Mass of the object is m = 490 g spring streches by the position is x = 39.0cm Amplitude of the oscillations is A = 15.5 cm Restoring force acting on the spring is F = k x                  k = 0.49kg * 9.8 m/s^2 / 0.39 m                      = 12.3 N /m Angular frequency = k / m                                      = 12.3 N / m / 0.490 kg                                     = 5.01 rad /s The position at the moment after t = 84.4 s is                 x = A cos t              = 0.155m cos ( 5.01 rad /s * 84.4 s)              = 0.070m                          = 7.01 cm b ) During this complete oscillation it covers the distance 4 A                   2 rad = 4 ( 0.155m)                                = 0.62                 1 rad = 0.62/ 2                          = 0.098m c ) If another hanging spring stretches by x = 39.5 cm           mass of the object is   m = 480g        Amplitude is A = 15.5 cm           Force F = kx                k = 0.480 kg * 9.8 m/s^2 / 0.395 m                   = 11.9 N /m Angular frequency = k / m                                      = 11.9 N / m / 0.480 kg                                     = 4.98 rad /s                                      = 11.9 N / m / 0.480 kg                                     = 4.98 rad /s The position at the moment after t = 84.4 s is                 x = A cos t              = 0.155m cos ( 4.98 rad /s * 84.4 s)              = 0.076m                          = 7.06 cm d ) During this complete oscillation it covers the distance 4 A                   2 rad = 4 ( 0.155m)                                = 0.62                 1 rad = 0.62/ 2                          = 0.098m                 x = A cos t              = 0.155m cos ( 4.98 rad /s * 84.4 s)              = 0.076m                          = 7.06 cm d ) During this complete oscillation it covers the distance 4 A                   2 rad = 4 ( 0.155m)                                = 0.62                 1 rad = 0.62/ 2                          = 0.098m                   2 rad = 4 ( 0.155m)                                = 0.62                 1 rad = 0.62/ 2                          = 0.098m

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