A 0.25-kg mass is hanging from a spring with spring constant 15 Nm. Then the mas
ID: 1466310 • Letter: A
Question
A 0.25-kg mass is hanging from a spring with spring constant 15 Nm. Then the mass is displaced from the equilibrium by 2.4 cm and let go. What is the resulting angular frequency of the oscillation? (use rad/s as units) A 0.69-kg mass is hanging from a spring with spring constant 17 N/m. Then the mass is displaced from the equilibrium by 2.6 cm and let go. What is the resulting frequency of the oscillation? A 0.67-kg mass is hanging from a spring with spring constant 12 N/m. Then the mass is displaced from the equilibrium by 1.5 cm and let go. What is the resulting period of the oscillation? A harmonic oscillator consisting of a spring with spring constant 9.9 N/m and an unknown mass has a frequency of 0.39 Hz. What is the mass? A harmonic oscillator consisting of a spring with unknown spring constant and a mass of 0.48 kg has a period of 1.4 s. What is the spring constant (use units N/m)? A harmonic oscillator that is used in an introductory physics lab consists of a spring with spring constant 35 Him and a number of different masses that can get attached to the spring. In the last week of the semester, the teaching assistant notices that all measured periods are about 21% higher than in the beginning of the semester. The TA suspects that the spring has been overstretched and that this has resulted in a permanent change of the spring constant. What is the new spring constant? A simple harmonic oscillator consisting of a mass attached to a spring has a frequency of 2.68 Hz. If we now increase the mass by 18%, what is the new frequency of the oscillator? An ideal massless spring is hanging vertically. When a 0.58-kg mass is attached to the spring, the spring stretches by 4.2 cm before reaching a new equilibrium. Then the mass is displaced by 2.7 cm from the newequlibrium. What is the frequency of the resulting oscillation?Explanation / Answer
a)
k = 15 N/m and x = 2.4 cm and m=0.25kg
angular-frequency = sqrt(k/m) = sqrt(60) = 7.745 radian per second
b)
m = 0.69 kg and k = 17 N/m and x = 2.6 cm
frequency = (1/2pi)sqrt(k/m) = (7/44)(4.963) = 0.78 Hz
c)
m = 0.67kg and k = 12 N/m and x = 15 cm
period = (2pi)sqrt(m/k) = (44/7)(0.2362) = 1.484s
d)
k = 9.9 N/m and f = 0.39Hz
frequency = (1/2pi)sqrt(k/m)
=> 0.39 = (7/44)sqrt(k/m)
=> 2.45 = sqrt(k/m)
=> 6.0025 = 9.9/m
=> m = 1.65 kg
e)
m = 0.48kg and T = 1.4s
period = (2pi)sqrt(m/k)
=> 1.4*7/44 = sqrt(m/k)
=> 0.0495 = m/k
=> k =0.48/0.0495 = 9.69N/m
g)
T = (2pi)sqrt(m/k)
T1 = T, T2 = 1.21T and k1 = k
T1/T2 = sqrt(k2/k1)
=> 1/1.21 = sqrt(k2/k)
=>0.683 = k2/k
=>k2 = 0.683k
h)
frequency = (1/2pi)sqrt(k/m)
f2/f1 = sqrt(m1/m2)
=> f2 = 2.68*sqrt(1/1.18) = 2.68*0.9205 = 2.47 Hz
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