Help Please!!Any help will be appreciated!!! 1) An ideal massless spring is hang

Question

Help Please!!Any help will be appreciated!!!

1)An ideal massless spring is hanging vertically. When an unknown mass is attached to the spring, the spring stretches by 4.8 cm before reaching a new equilibrium. Then the mass is displaced by 1.3 cm from the new equlibrium. What is the frequency of the resulting oscillation?

2) A common problem inside orbiting space stations is monitoring astronauts weight, or better said mass. The apparent weightlessness does not allow the use of conventional scales (note that due to the close proximity to earth, the astronauts weight did not really change drastically.)

Explanation / Answer

from force balance

mass be m of the unknown object

m*g = k*x

=> m*9.8 = k*0.048

=> k/m = 204.16

frequncy is known to be dependent on this ratio only

? = 2*?*f = sqrt(k/m) sqrt means square root

for reference of this formula http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

? = sqrt(204.16) = 14.28s-1 = 2*?*f

f = 14.28/(2*?) = 2.2727 s-1 = 2.2727 Hz

2)

t for chair = T

t with astro = (1+1.20)T = 2.20T

we know t is inverse of frequency or we cn say t is inversely related to ?

also, ?= sqrt(k/m)=> ? = 2?/t sqrt means square root of

?Tnew/Told = sqrt(mnew/mold)

2.20 = sqrt(mnew/20)

2.20^2 = mnew/22

mnew = 106.48

m of chair = 20

thus mass of astronaut = 106.48 - 22 = 84.48 kg

3)

T = 2??(m/k)

1.4 = 2*3.14*?(m/k)

m/k = 0.04969

m = 0.04969*k

Total energy = 1/2*ka^2 = 1/2*k*0.041^2

When half way from equilibrium position, Total energy = KE + PE = 1/2*mv^2 + 1/2*kx^2

= 1/2*mv^2 + 1/2*k*(a/2)^2

= 1/2*m*v^2 + 1/2*k*0.041^2/4

Energy conservation: 1/2*k*0.041^2 = 1/2*m*v^2 + 1/2*k*0.041^2/4

Or, 1/2*mv^2 = 1/2*k*(3/4)*0.041^2

1/2*(0.04969*k)v^2 = 1/2*k*(3/4)*0.041^2

v = 0.1592 m/s

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