# Finding the period of a system in simple harmonic motion [closed]

A snowball of mass 100g is thrown with a velocity of 10 m/s horizontally on a box on a frictionless table. The box is connected to a spring which is attached to the wall. The box is of mass 1.0kg. The entire snowball is attached to the box and it is observed that the box is compressed 16 cm at most after the snowball hits the box.

a) explain why this starts the system to be in simple harmonic motion.

b) find the period for the oscillation of the system.

c) what happens too the period if the snowball falls off and the box is the only thing oscillating.

I think I got the first problem right, it is in simple harmonic motion because the force of the snowball is the only force acting on the system, so the acceleration is directly proportional to the displacement. a= -kx/m (by hooks law, newtons 2. and 3. law)

In the second problem I have calculated the final velocity by use of the conservation of momentum in perfectly inelastic collisions. But I can't figure out how to proceed from here. Since I think I have to find the spring constant, but I dont know how to find it.

All help is appreciated!

## closed as off-topic by John Rennie, Jon Custer, user108787, user36790, GertNov 6 '16 at 20:54

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You have the right approach in using conservation of momentum.

(a) the system results in oscillatory motion because the momentum of the snowball alters the momentum of the box upon collision, and the impulse sets the box-spring component out of equilibrium. The distance the box is moved out of equilibrium will tell you the restoring force of the spring, depending on the spring constant $k$, within the equation $F = -kx$

(b) the question tells you the displacement of the box after collision, $x = 16cm$.

conservation of momentum tells you that the momentum of the snowball after equals the momentum of the system after, so letting $p_1 =$ snowball momentum and $p_2 =$ box momentum, and $p = mv$

$p_1 = (m_1 + m_2)v$

we know $p_1 = (0.1kg)(10m/s) = 1kgm/s$

$m_1+m_2 = 1.1kg$

so the velocity of the system after collision is $v = \frac{p_1}{m_1+m_2} = 0.91m/s$

you know that all the force in the system comes from the snowball initially, so using your relation $ma = -kx$, you know the spring constant $k = -\frac{ma}{x}$

the question reduces to: you have a mass of 1.1kg which compresses a spring by 0.16m; what is the spring constant $k$? with a vertical spring under gravity, $a = -9.81m/s^2$ but what is the acceleration here? it can be calculated using the kinematic equation $v_f = v_i + 2ax$ where you know that at the point of maximum compression, $v_f = 0$. initially, velocity is $v_i = 0.91m/s$, and the system compresses at $x = .16m$

then the period of oscillation is given by $T = 2\pi \sqrt\frac{m}{k}$ where m is the system mass, and k is spring constant.

(c) if the snowball does not stick to the box, what changes to the system? the momentum of the system will now depend only on the mass of the box, so the momentum would be lower. We would expect the box not to compress the spring as much.

but simply looking at the equation for period $T$, we see the only variable is mass $m$. the spring constant $k$ is... a constant, so we can see that the period will change only according to the missing mass of the snowball no longer combined with the box mass.

$T_1 = 2\pi \sqrt\frac{(m_1+m_2)}{k}$

$T_2 = 2\pi \sqrt\frac{(m_2)}{k}$

where $m_2$ is the box mass.