QUESTION 1 A 0.51-kg mass is hanging from a spring with spring constant 13 N/m.

Question

QUESTION 1 A 0.51-kg mass is hanging from a spring with spring constant 13 N/m. Then the mass is displaced from the equilibrium by 2.1 cm and let go. What is the resulting angular frequency of the oscillation? (use rad/s as units)

QUESTION 2 A 0.82-kg mass is hanging from a spring with spring constant 10 N/m. Then the mass is displaced from the equilibrium by 1.7 cm and let go. What is the resulting frequency of the oscillation?

QUESTION 3 A 0.27-kg mass is hanging from a spring with spring constant 15 N/m. Then the mass is displaced from the equilibrium by 2.7 cm and let go. What is the resulting period of the oscillation?

QUESTION 4 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?

QUESTION 5 A harmonic oscillator consisting of a spring with unknown spring constant and a mass of 0.36 kg has a period of 0.7 s. What is the spring constant (use units N/m)?

QUESTION 6 A harmonic oscillator that is used in an introductory physics lab consists of a spring with spring constant 34 N/m 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 33% 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?

QUESTION 7 A simple harmonic oscillator consisting of a mass attached to a spring has a frequency of 2.94 Hz. If we now increase the mass by 30%, what is the new frequency of the oscillator?

QUESTION 8 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 new equlibrium. What is the frequency of the resulting oscillation?

Explanation / Answer

1)
Angular Frequency of Oscillation,   is given by
= sqrt(k/m)
= sqrt(13/0.51) rad/s
= 5.05 rad/s

2)
Frequency of the oscillation is given by ,
f = 1/(2*pi) * sqrt(k/m)
f = 1/(2 * 3.14 ) * sqrt(10/0.82)
f = 0.556 Hz

3)
Time period is given by , T = 1/f
T = 2*pi * sqrt(m/k)
T = 2*3.14 * sqrt(0.27/15)
T = 0.843 s

4)
k = 9.9 N/m
f = 0.39 Hz

2*pi*f = sqrt(k/m)
2*pi * 0.39 = sqrt(9.9/m)
Solving for m
m = 1.65 Kg

Please post seperate questions in seperate post

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