SATHEE: Simple Harmonic Motion 5 Question 3

Simple Harmonic Motion 5 Question 3

3. A simple pendulum of length $1 m$ is oscillating with an angular frequency $10 r a d / s$ . The support of the pendulum starts oscillating up and down with a small angular frequency of $1 r a d / s$ and an amplitude of $10^{- 2} m$ . The relative change in the angular frequency of the pendulum is best given by

(a) $1 r a d / s$

(b) $10^{- 5} r a d / s$

(d) $10^{- 1} r a d / s$

(2019 Main, 11 Jan II)

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Answer:

Correct Answer: 3. (c)

Solution:

We know that time period of a pendulum is given by

$T = 2 π \sqrt{\frac{l}{g}}$

So, angular frequency $ω = \frac{2 π}{T} = \sqrt{\frac{g}{l}}$

Now, differentiate both side w.r.t $g$

$\begin{aligned} ∴ \frac{d ω}{d g} & = \frac{1}{2 \sqrt{g} \sqrt{l}} \\ d ω & = \frac{d g}{2 \sqrt{g} \sqrt{l}} \end{aligned}$

By dividing Eq. (ii) by Eq. (i), we get

$\frac{d ω}{ω} = \frac{d g}{2 g}$

Or we can write

$\frac{Δ ω}{ω} = \frac{Δ g}{2 g}$

As $Δ g$ is due to oscillation of support.

$∴ Δ g = 2 ω^{2} A$

$(ω_{1} \to 1 r a d / s$ , support)

Putting value of $Δ g$ in Eq. (iii) we get

$\begin{aligned} \frac{Δ ω}{ω} = \frac{1}{2} \cdot \frac{2 ω_{1}^{2} A}{g} = \frac{ω_{1}^{2} A}{g}; (A = 10^{- 2} m^{2}) \\ \Rightarrow & \frac{Δ ω}{ω} = \frac{1 \times 10^{- 2}}{10} = 10^{- 3} r a d / s \end{aligned}$

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Simple Harmonic Motion 5 Question 3

Answer:

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Simple Harmonic Motion 5 Question 3

3. A simple pendulum of length 1m is oscillating with an angular frequency 10rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1rad/s and an amplitude of 10−2m. The relative change in the angular frequency of the pendulum is best given by

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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