4 Marks Questions

Question 14 Marks

A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.

Answer

When the car moving with uniform velocity,

$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$

$\Rightarrow4=2\pi\sqrt{\frac{\ell}{\text{g}}}\ ...(1)$

When the car makes accelerated motion, let the acceleration be a₀

$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}^2+\text{a}_0^2}}$

$\Rightarrow3.99=2\pi\sqrt{\frac{\ell}{\text{g}^2+\text{a}_0^2}}$

Now, $\frac{\text{T}}{\text{T}'}=\frac{4}{3.99}=\frac{\big(\text{g}^2+\text{a}_0^2\big)^{\frac{1}{4}}}{\sqrt{\text{g}}}$

Solving for ‘a₀’ we can get a₀ $=\frac{\text{g}\text{}}{10}\text{ms}^{-2}$

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Question 24 Marks

A simple pendulum of length I is suspended from the ceiling of a car moving with a speed v on a circular horizontal road of radius r.

Find the tension in the string when it is at rest with respect to the car.
Find the time period of small oscillation.

Answer

From the freebody diagram,

$\text{T}=\sqrt{(\text{mg})^2+\Big(\frac{\text{mv}^2}{\text{r}^2}\Big)}$

$=\text{m}\sqrt{\text{g}^2+\frac{\text{v}^4}{\text{r}^2}}=\text{ma}$ where a = acceleration $=\Big(\text{g}^2+\frac{\text{v}^4}{\text{r}^2}\Big)^{\frac{1}{2}}$

The time period of small accellations is given by,

$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}=2\pi\sqrt{\frac{\ell}{\Big(\text{g}^2+\frac{\text{v}^4}{\text{r}^2}\Big)^{\frac{1}{2}}}}$

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Question 34 Marks

The ear-ring of a lady shown in has a 3cm long light suspension wire.

Find the time period of small oscillations if the lady is standing on the ground.
The lady now sits in a merry-go-round moving at 4m/s in a circle of radius 2m. Find the time period of small oscillations of the ear-ring.

Answer

$\ell=3\text{cm}=0.03\text{m}.$

$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}=2\pi\sqrt{\frac{0.03}{9.8}}=0.34$ secound.

When the lady sets on the Merry-go-round the ear rings also experience centrepetal acceleration,

$\text{a}=\frac{\text{v}^2}{\text{r}}=\frac{\text{4}^2}{2}=8\text{m/s}^2$

Resultant Acceleration $\text{A}=\sqrt{\text{g}^2+\text{a}^2}=\sqrt{100+64}=12.8\text{m/s}^2$

Time period $\text{T}=2\pi\sqrt{\frac{\ell}{\text{A}}}=2\pi\sqrt{\frac{0.03}{12.8}}=0.30$ second.

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Question 44 Marks

A person goes to bed at sharp 10:00 pm every day. Is it an example of periodic motion? If yes, what is the time period? If no, why?

Answer

It is not motion at first place.

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