Determining Mass Using Spring Balance Oscillation

This problem requires us to calculate the mass of a body by analyzing its vertical oscillation frequency when attached to a spring balance. We will use concepts from Simple Harmonic Motion (SHM).

Calculating the Spring Constant (k)

First, we need to determine the spring constant ($k$) of the spring balance. The balance has a range of 0 to 200 N and a length of 10 cm. We assume the spring behaves linearly according to Hooke's Law ($F = kx$), where $F$ is the force and $x$ is the extension. The range implies that a force of 200 N causes an extension of 10 cm.

Given:

• Maximum Force ($F_{max}$) = 200 N
• Maximum Extension ($x_{max}$) = 10 cm = 0.1 m

Using Hooke's Law, we find the spring constant ($k$):

$k = \frac{F_{max}}{x_{max}}$

$k = \frac{200 \text{ N}}{0.1 \text{ m}}$

$k = 2000 \text{ N/m}$

Calculating the Mass (m)

The body oscillates vertically at a frequency ($f$) of 2 Hz. The formula relating frequency, spring constant, and mass in SHM is:

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

We need to find the mass ($m$). Let's rearrange the formula to solve for $m$.

1. Square both sides of the equation: $f^2 = \frac{1}{4\pi^2} \frac{k}{m}$
2. Isolate $m$: $m = \frac{k}{4\pi^2 f^2}$

Now, substitute the values we have:

• Spring Constant ($k$) = 2000 N/m
• Frequency ($f$) = 2 Hz

Plug these values into the rearranged formula:

$m = \frac{2000 \text{ N/m}}{4\pi^2 (2 \text{ Hz})^2}$

$m = \frac{2000}{4\pi^2 \times 4}$

$m = \frac{2000}{16\pi^2}$

To get a numerical answer, we can use the approximation $\pi \approx 3.14159$. Then $\pi^2 \approx 9.8696$.

$m \approx \frac{2000}{16 \times 9.8696}$

$m \approx \frac{2000}{157.9136}$

$m \approx 12.665 \text{ kg}$

Final Answer Comparison

The calculated mass is approximately 12.665 kg. Comparing this value to the given options:

• 22.5 kg
• 12.5 kg
• 37 kg
• 45 kg

The closest option to our calculated value is 12.5 kg.

Therefore, the mass of the body is nearly 12.5 kg.