1) A 5 kg mass attached to a spring is pulled back 0.3 m and released from a horizontal frictionless table. The system has a frequency of 0.7 Hz. 2) Using the given frequency and mass, the spring constant is calculated to be 96.7 N/m. 3) The maximum velocity of the mass is calculated to be 1.32 m/s using the spring constant, mass, and initial displacement. 4) If the mass was larger but the spring constant remained the same, the frequency would decrease because increasing the mass in the denominator of the frequency equation would lower the value of the square root term.

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.!