What is the ratio of masses of the bob in a spring experiment?

[desmond iking]

Homework Statement

i have attached the sample ans here.
hand written photo is my working.

Homework Equations

The Attempt at a Solution

since the question doesn't same string . so i assume it's a same spring. so i would get t is directly propotional to sqrt root (m). is my concept wrong?

 

[Simon Bridge]

The question states that the string is elastic - so what effect would different masses of bob have on the string?
Note: the question asks you to explain the effect - your working is maths only, an explanation would have words in it.

 

[desmond iking]

The question states that the string is elastic - so what effect would different masses of bob have on the string?
Note: the question asks you to explain the effect - your working is maths only, an explanation would have words in it.

post deleted~

 

[Simon Bridge]

I don't know - I cannot follow it.
That is why I'm asking about your reasoning.

 

[desmond iking]

I don't know - I cannot follow it.
That is why I'm asking about your reasoning.

hi, i am just asking the maths part.the spring is elastic, so the extension of spring would be different for both cases. why can't I use the above working ?since this is a spring-mass system. so T=2pi(m/k) 1/2... taking 2pi(k)^-0.5 as constant. T is directly propotional to m

 

[Simon Bridge]

The x in F=kx is the extension ... that is the change in length of the spring.
The pendulum string has an unstretched length as well.
Did you take that into account?

How did you deal with the string changing length during the swing?

 

[desmond iking]

The x in F=kx is the extension ... that is the change in length of the spring.
The pendulum string has an unstretched length as well.
Did you take that into account?

How did you deal with the string changing length during the swing?

in other words, i let (m/k)=(l/g) since mg=kl. so i change the ratio of (l/g) into
(m/k)... so i get T is directly propotional to m

 

[ehild]

You assumed the bob vibrating vertically (first picture), while the problem speaks about a pendulum, the bob moving along an arc of circle about the equilibrium point (second picture). As it is an elastic string, the bob can move in a quite complicated way. If it is initially in rest, and you pull a little bit vertically, it performs vertical SHM with ω=√(k/m). That is not pendulum motion.
If you kick the bob horizontally, it moves like a pendulum in the vertical plane. In case the deviation from vertical is very small, the length of the string does not change appreciably. For that motion, ω=√(g/l), the same as that of a "normal" pendulum. You need to use that formula, keeping in mind, that the length of the pendulum is not proportional to m. The string stretches because of the weigth of the bob, and the change of the length is proportional to the mass.


ehild

 

[desmond iking]

i can understand it finally. thanks

 

[ehild]

You are welcome

ehild

 

[Simon Bridge]

It's a common wrinkle in teaching labs where we deliberately give students a slightly stretchy string and get them to investigate the effect of changing the mass on the period - though, the size of the bob (as weights are added) is usually a bigger effect.

Students are startled to discover a mass dependence - it's character building ;)

Related - pendulum on a spring.
https://www.physicsforums.com/showthread.php?t=538269

The only niggle I have at the back of my mind is that this is the sort of question were the person setting it may be making some unspoken simplifications... kinda relying on the limited knowledge of the class to narrow the possible options.