a 0.50-kg mass hangs from a spring attached to the top of an elevator. the sprin

Question

a 0.50-kg mass hangs from a spring attached to the top of an elevator. the spring constant of the spring is 19.6 N/m. the elevator is moving upwards at a constant speed of 2.50 m/s, and the spring is stationary relative to the elevator. suddenly the elevator stops and the spring starts to oscillate.

(i) what is the period of the spring motion?

(ii) what is the amplitude of the spring motion?

(iii) a very light pebble sits on the top of the mass. does the pebble ever lose contact (Fn=0) with the mass? if so, find the postion where the pebble loses contact. if not, show why not.

Explanation / Answer

mass hanging from the spring is m = 0.50 kg

the spring constant of the spring is k = 19.6 N/m

speed of elevator is v = 2.50 m/s

(i)the angular frequency of the spring is

w = (k/m)^1/2

or 2pi x f = (k/m)^1/2

or 2pi x (1/T) = (k/m)^1/2

or T = 2pi x (m/k)^1/2

where T is period of spring motion

(ii)let A be the amplitude of the spring motion

we know that

v = A x w

or A = (v/w)

(iii)let the pebble lose contact at distance d

we know that

(1/2)mv^2 = (1/2)kd^2

or d^2 = (m/k) x v^2

or d = (m/k)^1/2 x v

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