SATHEE: Simple Harmonic Motion 3 Question 1

Simple Harmonic Motion 3 Question 1

1. The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 0.1 t} \cos (10 π t + φ)$ .

Here, $t$ is in seconds.

The time taken for its amplitude of vibration to drop to half of its initial value is close to

(2019 Main, 10 April I)

(a) $27 s$

(b) $13 s$

(d) $7 s$

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Answer:

Correct Answer: 1. (d)

Solution:

Given, displacement is

$x (t) = e^{- 0.1 t} \cos (10 π t + φ)$

Here, amplitude of the oscillator is

$A = e^{- 0.1 t}$

Let it takes $t$ seconds for amplitude to be dropped by half.

$\begin{aligned} t = 0 \Rightarrow A = 1 \\ t = t \Rightarrow A^{'} = \frac{A}{2} = \frac{1}{2} \end{aligned}$

[from Eq. (i)]

So, Eq. (i) can be written as

$\begin{aligned} e^{- 0.1 t} & = \frac{1}{2} \\ or e^{0.1 t} & = 2 \end{aligned}$

$\begin{aligned} or 0.1 t = \ln (2) \\ or \\ t = \frac{1}{0.1} \ln (2) = 10 \ln (2) \end{aligned}$

Now, $\ln (2) = 0.693$

$\begin{aligned} ∴ & t & = 10 \times 0.693 = 6.93 s \\ or & t & \approx 7 s \end{aligned}$

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अगला

Simple Harmonic Motion 3 Question 1

1. The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 0.1 t} \cos (10 π t + φ)$ .

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Simple Harmonic Motion 3 Question 1

1. The displacement of a damped harmonic oscillator is given by x(t)=e−0.1tcos⁡(10πt+φ).

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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1. The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 0.1 t} \cos (10 π t + φ)$ .