Chapter 23: Problem 2781
Complete the following sentence. Time period of oscillation of a simple pendulum is dependent on......... (A) Length of thread (B) initial phase (C) amplitude (D) mass of bob
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For a pendulum in damped oscillation, with a bob of mass \(\mathrm{m}\) and radius \(\mathrm{r}\), with a string of length \(\ell\) What is the time period? (A) \(\mathrm{T}=2 \pi \sqrt{(\ell / \mathrm{g})}\) (B) T depends on \(\mathrm{m}\) (C) \(\mathrm{T}>2 \pi \sqrt{(\ell / \mathrm{g})}\) (D) \(\mathrm{T}<2 \pi \sqrt{(\ell / \mathrm{g})}\)
In a damped oscillation with damping constant \(b\). The time taken for amplitude of oscillation to drop to half what is its initial value? (A) \((\mathrm{b} / \mathrm{m}) \ln 2\) (B) (b / \(2 \mathrm{~m}) \ln 2\) (C) \((\mathrm{m} / \mathrm{b}) \ln 2\) (D) \((2 \mathrm{~m} / \mathrm{b}) \ln 2\)
If the time period of undamped oscillation is \(\mathrm{T}\) and that of damped oscillation is \(\mathrm{T}^{\prime}\), then what is the relation between \(\mathrm{T} \& \mathrm{~T}^{\prime}\) (A) \(\mathrm{T}^{\prime}<\mathrm{T}\) (B) nothing can be said, unless the drag force constant is known. (C) \(\mathrm{T}^{\prime}=\mathrm{T}\) (D) \(\mathrm{T}^{\prime}>\mathrm{T}\)
In a damped oscillation with damping constant \(b\). The time taken for its mechanical energy to drop to half. What is its value? (A) \((\mathrm{b} / \mathrm{m}) \ln 2\) (B) \((\mathrm{b} / 2 \mathrm{~m}) \ln 2\) (C) \((\mathrm{m} / \mathrm{b}) \ln 2\) (D) \((2 \mathrm{~m} / \mathrm{b}) \ln 2\)
What is the equation for a damped oscillator, where \(\mathrm{k}\) and \(\mathrm{b}\) are constants and \(\mathrm{x}\) is displacement. (A) $\left\\{\left(\mathrm{md}^{2} \mathrm{x}\right) /\left(\mathrm{dt}^{2}\right)\right\\}+\mathrm{kx}+\\{(\mathrm{dbx}) /(\mathrm{dt})\\}=0$ (B) $\left\\{\left(\mathrm{md}^{2} \mathrm{x}\right) /\left(\mathrm{dt}^{2}\right)\right\\} \times \mathrm{kx}=\\{(\mathrm{dbx}) /(\mathrm{d} \mathrm{t})\\}$ (C) $m\left\\{\left(d^{2} x\right) /\left(d t^{2}\right)\right\\}=k x+\\{(d b x) /(d t)\\}$ (D) $\left\\{\left(\mathrm{md}^{2} \mathrm{x}\right) /\left(\mathrm{dt}^{2}\right)\right\\}-\mathrm{kx}=\mathrm{b}\\{(\mathrm{d} \mathrm{x}) /(\mathrm{dt})\\}$
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