Question

If a mass-spring system has a frequency of 9.7 Hertz, and a spring constant of 242 N/m, how much mass is on the spring? Answer to the nearest gram.

Answer 1

Approximately
`65\; {\rm g}` (assuming that the spring is horizontal, and that mass of the spring is negligible.)

Step-by-step explanation:

In a simple harmonic motion, assuming that the mass is at position
`0` when time is
`0`, the position
`x` of the mass (the oscillator) at time
`t` would be:

`x = A\, \sin(\omega\, t)`, where:

• 
  `A` is the amplitude of the motion, and
• 
  `\omega` is the angular velocity of the motion.

Note that
`\omega = 2\, \pi\, f` where
`f` is the frequency of the motion. It is given that
`f = 9.7\; {\rm s^(-1)}` in this question.

Differentiate
`x = A\, \sin(\omega\, t)` with respect to time
`t` to find the velocity of the oscillator at time
`t`:

`v = A\, \omega \, \cos(\omega\, t)`.

Differentiate again to find the acceleration of the oscillator:

`a = (-A\, \omega^(2))\, \sin(\omega\, t)`.

Let
`m` denote the mass of the oscillator and assume that mass of the spring is negligible. Assuming that the spring is horizontal, since all other force on the spring are balanced, the net force on the mass would be equal to the restoring force from the spring. By Newton's Laws of Motion, this net force would be:

`F = m\, a = m\, (-A\, \omega^(2))\, \sin(\omega\, t)`.

Divide the restoring force (which is equal to the net force) by displacement to find an expression for the spring constant:

`\begin{aligned}k &= -(F)/(x) \\ &= -(m\, (-A\, \omega^(2))\, \sin(\omega\, t))/(A\, \sin(\omega\, t)) \\ &= m\, \omega^(2)\end{aligned}`.

Since
`\omega = 2\, \pi\, f`, this equation becomes:

`k = m\, \omega^(2) = (2\, \pi\, f)^(2)\, m`.

Rearrange this equation to find mass
`m`:

`\begin{aligned}m &= (k)/((2\, \pi\, f)^(2)) \\ &= \frac{(242\; {\rm N\cdot m^(-1)})}{(2\, \pi\, (9.7\; {\rm s^(-1)}))^(2)} \\ &\approx 0.0651\; {\rm N\cdot m^(-1)\cdot s^(2)}\end{aligned}`.

Note that
`1\; {\rm N} = 1\; {\rm kg\cdot m\cdot s^(-2)}`. Hence:

`\begin{aligned}m &\approx 0.0651\; {\rm N\cdot m^(-1)\cdot s^(2)} \\ &= 0.065\; ({\rm kg\cdot m\cdot s^(-2)})\, ({\rm m^(-1)})\, ({\rm s^(2)}) \\ &= 0.0651\; {\rm kg\end{aligned}`.

Apply unit conversion and round to the nearest gram:

`m\approx 0.0651\; {\rm kg} \approx 65\; {\rm g}`.