# Mass hangs in equilibrium on a spring (oscillation)(MCQ)

Tags:
1. Feb 20, 2016

### slymme

1. The problem statement, all variables and given/known data
A mass M hangs in equilibrium on a spring. M is made to oscillate about the equilibrium position by pulling it down 10 cm and releasing it. The time for M to travel back to the equilibrium position for the first time is 0.5s. Which line, A to D, is correct for these oscillations?
A) Amplitude(cm) 10 , Period(s) 1.0
B) Amplitude(cm) 10 , Period(s) 2.0
C) Amplitude(cm) 20, Period(s) 2.0
D) Amplitude(cm) 20, Period(s) 1.0

2. Relevant equations
I attempted to solve this without the use of equations, but I guess these could be relevant:
T = t/n
T = 1/f
a = -w*2 x

3. The attempt at a solution
I may have solved the problem without actually knowing it, but its good to make sure. My first guess is when I read "by pulling it down 10 cm and releasing it". I guessed this to be the amplitude since it was released from the extreme position (or atleast I think so), so that leaves me with anwser A or B. Since M was released from the extreme position, It means it would take it 2x0.5 seconds to get back and give us the time period.Therefore, I went for answer A.
Some confirmation would be great. Forgive me if I may have made some mistakes posting this or related - this is my first post on this site :)

2. Feb 20, 2016

### ehild

0.5 s is needed to get to the equilibrium position from one of the extreme positions. But there are two extremes during one period. How much time is needed to go from one extreme to the opposite extreme?

3. Feb 20, 2016

### slymme

1 second? so does that mean the time period is 2 seconds and thus the answer being B?

4. Feb 20, 2016

### ehild

Yes

5. Feb 20, 2016

### slymme

I thank you good sir!

6. Feb 20, 2016

### ehild

You are welcome