Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A pendulum with the time period of 1 s is losing energy due to damping. At a certain time, its energy is 45 J. If after completing 15 oscillations its energy has become 15 J, then its damping constant (in s$$^{-1}$$) will be
We know that for a damped oscillator the displacement (or amplitude) decreases exponentially with time. Mathematically, if the initial amplitude is $$A_0$$ then at time $$t$$ we have
$$A(t)=A_0 e^{-\beta t}$$
where $$\beta$$ is the damping constant (its SI unit is $$s^{-1}$$). The total mechanical energy of the oscillator is proportional to the square of its amplitude, so
$$E(t)\propto [A(t)]^{2}$$
Substituting the expression for $$A(t)$$, the energy at time $$t$$ becomes
$$E(t)=E_0 e^{-2\beta t}$$
Here $$E_0$$ is the energy at $$t=0$$. Let us now insert the numerical data given in the problem.
At $$t=0$$ the energy is $$E_0=45\text{ J}$$. After the pendulum completes 15 oscillations its energy has dropped to $$E=15\text{ J}$$.
The time taken for one oscillation is the time period $$T=1\text{ s}$$, so the time needed to complete 15 oscillations is
$$t = 15T = 15 \times 1\text{ s} = 15\text{ s}.$$
Now we substitute $$E=15\text{ J}$$, $$E_0=45\text{ J}$$, and $$t=15\text{ s}$$ into the energy-decay equation $$E = E_0 e^{-2\beta t}$$:
$$15 = 45\, e^{-2\beta (15)}.$$
First divide both sides by 45 to isolate the exponential term:
$$\frac{15}{45} = e^{-30\beta}.$$
Simplifying the fraction on the left gives
$$\frac{1}{3} = e^{-30\beta}.$$
To solve for $$\beta$$ take the natural logarithm (base $$e$$) of both sides. Recall the logarithmic identity $$\ln(e^{x}) = x$$.
$$\ln\!\left(\frac{1}{3}\right) = \ln\!\left(e^{-30\beta}\right) = -30\beta.$$
Hence
$$\beta = -\,\frac{1}{30}\,\ln\!\left(\frac{1}{3}\right).$$
We can simplify further because $$\ln\!\left(\frac{1}{3}\right) = -\ln 3$$. Substituting this identity gives
$$\beta = -\frac{1}{30}\,(-\ln 3) = \frac{\ln 3}{30}.$$
Thus the damping constant is
$$\beta = \frac{1}{30}\,\ln 3\;\; \text{s}^{-1}.$$
This matches Option C.
Hence, the correct answer is Option C.
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.