Question

14. A mass m = 0.25 kg that is attached to a spring with spring constant k = 100 N/m is stretched 10 cm to the right of its equilibrium position. After a time t = 1 s, the mass is how far from the equilibrium position? ((+) equals to the right. (-) to the left.)

a) +4.08 cm b) +2.97 cm c) -9.63 cm d) -1.28 cm e) -6.67 cm

15. A simple pendulum has a period of 3 seconds on Earth. If it is taken to Planet X, which has the same mass as Earth but a radius 0.5 the size of Earth’s radius, the new period will be:

a) 2.40 s b) 2.10 s c) 1.80 s d) 2.00 s e) 1.50 s

Answer 1

14) The equation is given as

x (t) = Acos(wt)

w = sqrt (k/m)

w = sqrt (100 / 0.25) = 20 rad/sec

so,

x(1) = 10 cos (20*1)

x = 4.08 cm

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15) we can find the value of g

g = GM/r²

g = 6.67e-11 * 5.67e24 / (0.5 * 6370000)²

g = 37.25 m/s²

Now, we need length

we will use time period on earth to find L

T = 2*Pi*L/g

putting T = 3 and g = 9.8

L = 2.234 m

so,

T ( on X) = 1.5 sec