Question

A block of mass 100g attached to a spring of spring constant=100N/m lies on a frictionless floor at a distance of 5cm from the opp. wall. The block is moved to compress the string by 10cm and then released. Find the time period of the oscillations -

1. 0.2s

2. 0.1s

3. 0.15s

4. 0.132s

Answer 1

Answer on Question #40688 – Physics – Mechanics | Kinematics | Dynamics

A block of mass 100g attached to a spring of spring constant=100N/m lies on a frictionless floor at a distance of 5cm from the opp. wall. The block is moved to compress the string by 10cm and then released. Find the time period of the oscillations -

1. 0.2s

2. 0.1s

3. 0.15s

4. 0.132s

Solution:

$$\begin{array}{l} \mathrm{m} = 0.1\,\mathrm{kg} - \text{mass of the block}; \\ \mathrm{k} = 100\,\frac{\mathrm{N}}{\mathrm{m}} - \text{spring constant}; \\ \mathrm{A}_1 = 0.1\,\mathrm{m} - \text{amplitude of the oscillations on the first half of the period}; \\ \mathrm{A}_2 = 0.05\,\mathrm{m} - \text{amplitude of the oscillations on the first half of the period}; \end{array}$$

The period is the time for a vibrating object to make one complete cycle of vibration. The variables that effect the period of a spring-mass system are the mass and the spring constant. The equation that relates these variables resembles the equation for the period of a pendulum. The equation is

$$\mathrm{T} = 2\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \quad (1)$$

Since the amplitude of the oscillations on the first half of the period is twice $\left(\frac{A_1}{A_2} = \frac{0.1\,\mathrm{m}}{0.05\,\mathrm{m}} = 2\right)$ the amplitude of the oscillations in the second half of the period (block hits the wall and bounces from it).

Thus, block passes first part of the distance with time $\mathrm{T}_1 = \frac{\mathrm{T}}{2}$ (half of the period) and second part of the distance with time $\mathrm{T}_2 = \frac{\mathrm{T}_1}{2} = \frac{\mathrm{T}}{4} \Rightarrow$

$$\mathrm{T}_{\text{complete}} = \mathrm{T}_1 + \mathrm{T}_2 = \frac{\mathrm{T}}{2} + \frac{\mathrm{T}}{4} = \frac{3}{4}\,\mathrm{T} \quad (2)$$

(1)in(2):

$$\mathrm{T}_{\text{complete}} = 2\pi \frac{3}{4} \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} = \frac{3\pi}{2} \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} = \frac{3\pi}{2} \sqrt{\frac{0.1\,\mathrm{kg}}{100\,\frac{\mathrm{N}}{\mathrm{m}}}} = 0.15\,\mathrm{s}$$

Answer: 3) 0.15s