Question 14 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
View full question & answer→It is not motion at first place.
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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,
$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}=2\pi\sqrt{\frac{\ell}{\Big(\text{g}^2+\frac{\text{v}^4}{\text{r}^2}\Big)^{\frac{1}{2}}}}$
View full question & 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,
$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}=2\pi\sqrt{\frac{\ell}{\Big(\text{g}^2+\frac{\text{v}^4}{\text{r}^2}\Big)^{\frac{1}{2}}}}$
Question 34 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
View full question & 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 $\text{a}_0 \text{T}=2\pi\sqrt{\frac{\ell}{\text{g}^2+\text{a}_0^2}}$
$\Rightarrow3.99=2\pi\sqrt{\frac{\ell}{\text{g}^2+\text{a}_0^2}}$
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 ‘$\text{a}_0$’ we can get $\text{a}_0=\frac{\text{g}\text{}}{10}\text{ms}^{-2}$
$\Rightarrow4=2\pi\sqrt{\frac{\ell}{\text{g}}}\ ...(1)$
When the car makes accelerated motion, let the acceleration be $\text{a}_0 \text{T}=2\pi\sqrt{\frac{\ell}{\text{g}^2+\text{a}_0^2}}$
$\Rightarrow3.99=2\pi\sqrt{\frac{\ell}{\text{g}^2+\text{a}_0^2}}$
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 ‘$\text{a}_0$’ we can get $\text{a}_0=\frac{\text{g}\text{}}{10}\text{ms}^{-2}$
Question 44 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
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.
View full question & answer→
- $\ell=3\text{cm}=0.03\text{m}.$
- When the lady sets on the Merry-go-round the ear rings also experience centrepetal acceleration,
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.