Are These Physics Assignment Answers Correct?

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The discussion revolves around a physics assignment involving the analysis of standing waves in a vibrating rope and harmonic motion of a mass on a spring. The original poster seeks confirmation on their answers to various questions related to frequency, angular frequency, maximum speed, and forces acting on beads in the system. Participants emphasize the importance of showing work for each problem to facilitate accurate feedback. The conversation highlights the collaborative nature of learning physics, encouraging peer assistance to ensure understanding and correctness in problem-solving.
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Hi to evryone!
It is so nice to know that everyone is really helpful here.
I have got a physics assignment to hand in tomorrow, although I have done it I am not so sure, I just wanted to check if I have done it right or not. U c I didn't attend many lessons in the last couple of weeks, so wasn't to sure about the assignment.
I would be ever so grateful if someone could tackle these questions for me, then I can check whether I have approached them in the right way. THANKYOU

1. A light rope, fixed at each end, is vibrating in a SHM due to waves passing through it. The waves are resonant with the rope and a standing wave has been set up in the rope. There are 3 small beads A, B and C, each of mass 2.10 g.
As the standing wave vibrates inthe rope, the bead A moves a total distance of 720 mm, the bead B remains stationary, while the bead C moves a total distance of 348mm.
The waves are traveling through the rope, which is 1.48m long (between the points of attachment), with a speed of 12.0ms(minus 1)

a) what is the frequency of the waves traveling through the rope?
b) what is the angular frquency for the vibration of;
Bead A Bead C
c) as they vibrate in their SHM, what is the maximum speed of:
Bead A Bead C
d) as the rope vibrates in the standing wave, what is the maximum size of the force which acts on;
Bead A Bead B Bead C

2. A 925 g. mass is hung from a spring and allowed to come to equilbrium. A force of 6.0 N then pulls vertically down the mass. The force causes the spring to stretch by 180mm.
When the mass is released it is observed to execute a uniform single harmonic motion.
Determine:
a) the amplitude of the SHM
b) the acceleration of the mass when it is released
c) the spring constant for the spring
d) the angular frequency of the SHM
e) the maximum speed of the mass as it oscillates
f) the period of the SHM
 
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Saira, folks here won't do the work for you. Instead, show us what your solution is, and they can tell you if you got the answer right or can help point out any mistakes you made.
 


Hi there, I'm happy to help with your physics assignment questions. It sounds like you've put in a lot of effort and just want to make sure you're on the right track. Let's take a look at the questions together.

For question 1, we have a light rope that is vibrating due to resonant waves passing through it. We have three beads, A, B, and C, with different masses and different distances they move as the rope vibrates. We also know the length of the rope and the speed of the waves traveling through it.

a) To find the frequency, we can use the formula f = v/λ, where f is the frequency, v is the speed of the waves, and λ is the wavelength. In this case, we know v = 12.0 m/s and λ = 1.48 m. Plugging these values in, we get f = 12.0/1.48 = 8.11 Hz.

b) The angular frequency is given by ω = 2πf, where ω is the angular frequency and f is the frequency we just found. So for bead A, ω = 2π(8.11) = 51.0 rad/s. For bead C, ω = 2π(8.11) = 51.0 rad/s as well.

c) To find the maximum speed, we can use the formula v_max = ωA, where v_max is the maximum speed, ω is the angular frequency, and A is the amplitude of the SHM. For bead A, v_max = 51.0(0.72) = 36.7 m/s. For bead C, v_max = 51.0(0.348) = 17.7 m/s.

d) The maximum force acting on the beads can be found using the formula F_max = mω²A, where F_max is the maximum force, m is the mass of the bead, ω is the angular frequency, and A is the amplitude. For bead A, F_max = (0.00210)(51.0)²(0.72) = 5.63 N. For bead B, F_max = (0.00210)(51.0)²(0) = 0 N. For bead C, F_max = (0.00210)(51.0)
 
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