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

  • #1
5
0

Homework Statement


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

Homework 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

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 :)
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,914

Homework Statement


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

Homework 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

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 :)
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
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0
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?
1 second? so does that mean the time period is 2 seconds and thus the answer being B?
 
  • #4
ehild
Homework Helper
15,543
1,914
1 second? so does that mean the time period is 2 seconds and thus the answer being B?
Yes
 
  • #6
ehild
Homework Helper
15,543
1,914
You are welcome:oldsmile:
 

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