What's Wrong with My SHM Calculations?

In summary, the mass, m, vibrates with a frequency of 1.06 Hz when an additional 1.0 kg mass is added to it. However, when an equation is used to find the value of m, the answer is incorrect. Can anyone take me through the working out so I know where I was wrong?
  • #1
nonzee
2
0
I've scrolled through other SHO problems and can't really find what I need. I have a homework-test thing and it's only on revision from high school (just started university) and I can't even get it all right!

Q1) A mass m at the end of a spring vibrates with a frequency of 1.06 Hz. When an additional 1.0 kg mass is added to m, the frequency is 0.38 Hz. What is the value of m?

I used the equation f = 1/2pi * (squareroot k/m)

I eventually came up with the answer 1/0.1718 but I've been told this answer is wrong? Can anyone take me through the working out so I know where I was wrong?

Q2) A tuning fork vibrates at a frequency of 237 Hz and the tip of each prong moves 1.93 mm at either side of centre. Calculate the maximum speed of each prong (in m/s).

I thought the formula for this would have been velocity(max) = 2pi*f*A, with 0.000193m as A. I got this formula from my textbook and ended up with 0.29 m/s. However, once again, I was told I was wrong!

Q5) It takes a force of 70 N to compress the spring of a popgun 0.197 m to load a 0.13 kg ball. With what speed will the ball leave the gun?

I'm dumbfounded. Where do I start? F = -kx? v(max) = 2pi*f*A? Somehow I got 10.3966 but i can't understand my own working out (and it's wrong).

Q6) A 0.0127 kg bullet strikes a 0.36 kg block attached to a fixed horizontal spring whose spring constant is 2.62 x103 N/m and sets it into vibration with an amplitude of 12.4 cm.
What was the speed of the bullet?

They failed to mention it, but the bullet actually sticks to the block, too. Anyway I tried to use my good old v(max) = (sqrt(k/m))*A again but my answer, 10.3900, I was informed that it was wrong.

If anyone could help, it'd be fantastic! I really need to know how to do this stuff and my lecturer is always unavailable during my free times.
 
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  • #2
You just need to form the ratio first frequency [tex]f_1[/tex] to second frequency [tex]f_2[/tex] and then solve for m:

[tex]\frac{f_1}{f_2} = \sqrt{\frac{m + 1}{m}}[/tex]
 
  • #3
Maybe you should stop looking through your book for formulas that seem to work and should instead start thinking about what each question is asking.

In simple harmonic motion, when is the the velocity greatest? When the forks are fully displaced? When they're not displaced at all? Somewhere inbetween? Once you know the position, how do you calculate the velocity at that position?

When the spring is fully compressed, it has a certain amount of energy stored. When the spring extends, all that stored energy is going to be converted to kinetic energy. So what does this mean for your problem?

Once again, energy! The bullet starts off with some energy that's all transferred to make the block oscillate.

cookiemonster
 
  • #4
Thankyou very much for your help :smile:
I always did have a weakness for conceptualising but I've finally figured it out.
And sorry for putting it in the wrong place! I'll know next time
 
  • #5
Originally posted by cookiemonster:
Maybe you should stop looking through your book for formulas that seem to work and should instead start thinking about what each question is asking.

Cookiemonster- you're starting to sound like me!
 

Related to What's Wrong with My SHM Calculations?

What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion (SHM) is a type of periodic motion in which an object oscillates back and forth around a central equilibrium point in a straight line. It is characterized by a restoring force that is directly proportional to the displacement from the equilibrium point, making the motion repetitive and predictable.

What are the factors that affect SHM?

The factors that affect SHM include the mass of the object, the amplitude of the oscillation, and the stiffness of the restoring force. The frequency of SHM is determined by the mass and stiffness, while the amplitude depends on the initial conditions of the motion.

How is SHM related to Hooke's Law?

Hooke's Law states that the force needed to extend or compress a spring is directly proportional to the distance it is extended or compressed. This is also the case for SHM, where the restoring force is proportional to the displacement from the equilibrium position. Thus, SHM and Hooke's Law are closely related concepts.

What are some applications of SHM?

SHM has many practical applications in various fields such as engineering, physics, and biology. Some common examples include the motion of a pendulum, the vibrations of a guitar string, and the motion of a mass-spring system. SHM is also used in seismology to study earthquakes and in medicine to measure heart rate and blood pressure.

How can the period and frequency of SHM be calculated?

The period (T) of SHM can be calculated by dividing the time it takes for one complete oscillation by the number of oscillations. The frequency (f) can be calculated by dividing the number of oscillations by the time taken. The relationship between period and frequency is f = 1/T. Additionally, the period and frequency can be affected by the factors mentioned in the second question.

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