Question

An object of mass 250 g is connected to two springs, both on the same side of the object, so that the springs are parallel to each other. The springs are the same length when unstretched; one has a spring constant of 75 N/m, and the other has a spring constant of 50 N/m. The object moves without friction on a horizontal surface. If the object is pulled to one side and released, what is the frequency of the resulting simple harmonic motion?

Answer 1

Approximately
`3.56\; {\rm Hz}`.

Step-by-step explanation:

Under the assumptions of this question, the object will be part of a spring-mass system and will be in a simple harmonic motion. To find the frequency of this motion, start by finding the effective spring constant. After that, derive the equation for the frequency of this spring-mass system. Substitute in the effective spring constant and the mass of the object to find the frequency of the motion.

Assume that two ideal springs of equal length are connected in parallel. The two springs would behave like one ideal spring with the spring constant equal to the sum of that of the two individual springs.

In this question, the spring constant of the two individual springs are
`75\; {\rm N\cdot m^(-1)}` and
`50\; {\rm N\cdot m^(-1)}`, respectively. When the two springs are connected as described, the effect of the two separate springs would be equivalent to that of one ideal spring with spring constant
`(75 + 50)\; {\rm N\cdot m^(-1)} = 125\; {\rm N\cdot m^(-1)}`.

Assume that the mass of the two springs are negligible. The expression for the frequency of this motion can be derived as follows:

In a frictionless horizontal spring-mass system, the net force on the moving object is equal to the restoring force from the spring. If the object is at a distance of
`x` from the equilibrium position, this restoring force would be
`(-k\, x)` where
`k` is the spring constant. (Negative since this force is opposite in direction to displacement.)

Divide this net force by mass to obtain the acceleration
`a` of the moving object:

`\begin{aligned} a &= \frac{(\text{net force})}{(\text{mass})} = (-k\, x)/(m) \end{aligned}`.

Assume that the moving object started at a distance of
`A` (amplitude) from the equilibrium position at time
`t = 0`.

Let
`f` denote the frequency of the motion. Since this object would be in a simple harmonic motion, position at time
`t` would be
`x = A\, \cos((2\, \pi\, f)\, t)`. The acceleration at time
`t\!` would be
`a = -A\, (2\, \pi\, f)^(2) \, \cos((2\, \pi\, f)\, t)`.

Substitute the expressions for
`x` and for
`a` into the equation
`\begin{aligned} a = (-k\, x) / m\end{aligned}` to obtain:

`\displaystyle -A\, (2\, \pi\, f)^(2) \, \cos((2\, \pi\, f)\, t) = (-k\, A\, \cos((2\, \pi\, f)\, t))/(m)`.

Simplify to obtain an expression for the frequency
`f` of the motion given the spring constant
`k` and mass
`m`:

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

`\displaystyle 2\, \pi\, f = \sqrt{(k)/(m)}`.

`\displaystyle f = (1)/(2\, \pi)\, \sqrt{(k)/(m)}`.

Ensure that mass is measured in the standard unit of kilograms:
`m = 250\; {\rm g} = 0.250\; {\rm kg}`. Substitute the value of mass and the effective spring constant
`k = 125\; {\rm N\cdot m^(-1)}` into the expression for frequency
`f` and evaluate:

`\begin{aligned} f &= (1)/(2\, \pi)\, \sqrt{(k)/(m)} \\ &= (1)/(2\, \pi)\, \sqrt{\frac{125\; {\rm N\cdot m^(-1)}}{0.250\; {\rm kg}}} \\ &= (1)/(2\, \pi)\, \sqrt{\frac{125\; {\rm kg\cdot s^(-2)}}{0.250\; {\rm kg}}} \\ &\approx 3.56\; {\rm s^(-1)} \\ &\approx 3.56\; {\rm Hz}\end{aligned}`.

In other words, the frequency of this motion would be approximately
`3.56\; {\rm Hz}`.