Question

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10cm. The maximum tension in the string is found to be \(\frac{x}{40}\) N. The value of x is___.

Answer 1

Correct Answer: 99

Given a simple pendulum, we need to find the value of \(x\) in the equation for the maximum tension \(\frac{x}{40}\) N. The pendulum length \(L=100\) cm = 1 m, mass \(m=250\) g = 0.25 kg, and amplitude \(A=10\) cm = 0.1 m. The formula for the tension in the pendulum string when it reaches the maximum amplitude is:

\[ T_{\text{max}} = m(g + \frac{v^2}{L}) \]

At maximum amplitude, the speed \(v\) is maximum at the mean position, so \(v^2 = A^2(\omega^2)\), where \(\omega = \sqrt{\frac{g}{L}}\).

Thus, \(v = A \cdot \omega\), and \(v^2 = A^2 \cdot \frac{g}{L}\).

Now substituting back:

\[ T_{\text{max}} = m \left(g + \frac{A^2 \cdot \frac{g}{L}}{L}\right) = mg + m \cdot \frac{A^2 \cdot g}{L^2} \]

Substitute the known values:

\[ T_{\text{max}} = 0.25 \times 9.8 + 0.25 \cdot \frac{(0.1)^2 \times 9.8}{(1)^2} \]

\[ = 2.45 + 0.25 \times \frac{0.01 \times 9.8}{1} = 2.45 + 0.0245 = 2.4745 \text{ N} \]

Since \(T_{\text{max}} = \frac{x}{40}\), we equate:

\[ \frac{x}{40} = 2.4745 \quad \Rightarrow \quad x = 2.4745 \times 40 = 98.98 \]

Verifying the range: The value \(x = 98.98\) falls within the expected range 99,99.