LO1

1. Horizontal Mass-Spring System!
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Jihyeon Kim!
15894141!
LM2!
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A 5.00 kg mass on a horizontal frictionless table is attached to a spring. The mass is pulled back
by 0.300 m and then released. The frequency of the system is 0.700 Hz.!
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a) Find the spring constant k of the spring.!
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b) Find the maximum velocity of the mass.!
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c) If the mass was the larger but the spring constant was the same as part a) would the
frequency of the system still be 0.700 Hz? Explain. !
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Solution!
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a) We are given that the mass is 5.00 kg and the frequency of the system is 0.700 Hz.!
To find the spring constant k, we first use the formula 𝑓= √(k/m)/(2π). (Pg.352 of textbook)!
𝑓 = √(k/m) / (2π)!
(2π)𝑓 = √(k/m)!
(2π𝑓)2 = k/m!
(2π𝑓)2m = k!
(2π × 0.700 Hz)2 (5.00 kg) = k!
∴ k = 96.7 N/m!
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b) The maximum velocity of the mass is given by v(t)= ωA. (Pg.349 of textbook)!
To use that formula, we need to calculate ω and A.!
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Finding ω:!

2. ω = √(k/m)!
= √96.7/5.00!
= 4.40 rad/s!
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To find A:!
x(0) = Acos(ωt + φ) = Acosφ !
v(0) = sinA(ωt + φ) = -ωAsinφ!
Combining these equations with trig identity we get!
A= √[(x2(0) + v2(0)/ω2]!
A= √(0.3002) + 0/4.402)!
= 0.300 m!
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v(t) = ωA!
= (4.40)(0.300)!
=1.32 m/s!
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c) The formula for frequency as mentioned above is 𝑓 = √(k/m) / (2π).!
When the mass increases, √(k/m) would decrease as the denominator is increasing.
Because √(k/m) has decreased, the frequency would be smaller.!