Question 7 of 14intermediate🔧 ApplyNumerical3 marks

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Correct Answer

The weight of the body is approximately 219 N (or 22.3 kg in terms of mass).

Exercise: EXERCISES | Q: 13.8 | (Page 17)
For More Understanding

Explanation

This problem combines the concept of spring constant determination from scale calibration with the time period formula of a spring-mass system executing SHM. The spring constant is first calculated from the given scale specifications, then used with the period formula to find the mass. The formula T = 2πm/k\sqrt{m/k} is derived from ω = k/m\sqrt{k/m} shown in Example 13.7 of the context.

Solution Steps

  1. Step 1: Calculate the spring constant (k) The scale reads 0 to 50 kg over a length of 20 cm (0.2 m). Maximum force = 50×9.850 \times 9.8 = 490 N Spring constant k = Force/extension = 490/0.2 = 2450 N/m

  2. Step 2: Use the time period formula for SHM From the formula T = 2πm/k\sqrt{m/k}, we get: m = kT²/(4π²) m = 2450 × (0.6)²/(4π²) m = 2450×0.362450 \times 0.36/(4×9.874 \times 9.87) m = 882/39.48 m = 22.34 kg

  3. Step 3: Calculate the weight Weight = mg = 22.34×9.822.34 \times 9.8 = 218.9 N ≈\approx 219 N

PreviousNext