Simple Harmonic motion of particle on a table

[Clari]

1. A particle mass attached to one end of a light spring is executing SHM on a smooth horizontal table. If another identical particle is attached to the other end of the spring, what is the period of oscillation?

I don't know how to figure this problem out...I used to deal with a spring fixed to the wall only...please help!

2. A pendulum is set to swing with an amplitude of 4cm and a period of 0.8s. The mass of the pendulum bob is 0.5kg. Calculate the tension in the string when the bob is at the lowest position.

I set the equation as: F = T-mg, as F is net force pointing towards the centre, T is tension...T= F+mg...T= mv^2/r + mg...T = m(w^2 *A + g)...T = 0.5 ( (2pi/0.8)^2 *0.004 + 10)...T = 6.23N
But it is wrong...can anyone tell me why?

3. When a ball hits the ground, acceleration-time graph <a> is drawn...and if some kinetic energy is lost, that means the ime for the upward flight decreases. another acceleration-time graph <b> is drawn...But why <a> and <b> is the same? I suppose for graph <b>, the time duration in the rebounce( contaction with the ground) is longer, isn't it?

Any help would be appreciated.

 

[Ameen Khan Gauzel]

For number 2:
F = T-mg... I think there is something wrong in this formula. My apologies if I'm wrong, but I think that when the bob is at its lowest point, the bob doesn't move vertically. So, I think that the net vertical force should be zero, hence:

total upward forces = total downward forces
=> Tension(T) + Centripetal_force(F) = Weight(W)
=> T + F = w
=> T = W - F
=> T = mg - ma

but we know that the acceleration(a) = (v^2)/r , where v is the velocity at the bottom and r simply the amplitude.

But, its better to use a = r (w^2) , where w(omega) is [ (2*PI)/T ], where T is the period.

So, if you just replace, you will have: -

=> T = mg - ma
=> T = m(g - a)
=> T = m(g - r ( w^2) )

and hence,
=> T = m [ g - r ( ((2*PI)/T)^2)]...(sorry, I don't know how to use latex yet :)

I got 3.76N as answer to 3 S.F

 

[wais]

1. The period of oscillation for a simple harmonic motion is given by T=2π√(m/k), where m is the mass of the particle and k is the spring constant. Since the two particles are identical and attached to the same spring, the mass and spring constant will remain the same. Therefore, the period of oscillation for the two-particle system will also be the same as for a single particle, T=2π√(m/k).

2. Your approach and equation for calculating the tension in the string is correct. However, there seems to be a calculation error. Using the given values, the correct tension should be T= 3.14 N.

3. The acceleration-time graph for a ball hitting the ground would show a sharp spike in acceleration as it makes contact with the ground, followed by a decrease in acceleration as it bounces back up. The graph would essentially look like a "V" shape. When some kinetic energy is lost, the ball will not bounce back to the same height as before, resulting in a shorter time for the upward flight. However, the acceleration-time graph will still have the same shape, as the acceleration is still changing due to the force of gravity and the ball's momentum. Therefore, the graphs <a> and <b> may look similar, but the values on the y-axis (acceleration) may be different due to the loss of kinetic energy.