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This is one of our many quizzes on Period and frequency. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Period and frequency Quiz 2

physics

The alarm produced by an industrial siren has a wavelength of $2.8\text{ m}$. If this alarm is propagating in air with a speed of $250\text{ m}{{\text{s}}^{-1}}$, evaluate its period. 

Recall the expressions for frequency as well as time period of oscillations. 

  $\text{Frequency is given by} \\ f=\dfrac{v}{\lambda } \\ \Rightarrow f=\dfrac{250}{2.8} \\ \Rightarrow f=89.2\text{ }Hz \\ \text{Now, time period is given by} \\ T=\dfrac{1}{f} \\ \Rightarrow T=\dfrac{1}{89.2} \\ \therefore T=0.01\text{ }\sec$ 

A signal is given by $x(t)=5\sin (25\pi t+0)$. Deduce an expression for the magnitude of frequency of this signal when the period is one-fourth of its actual magnitude in relation with the actual frequency. 

Compare the given equation with the general form of equation of wave. Connect the relation between time period and angular frequency.

  $\text{Time period is given by} \\ T=\dfrac{2\pi }{\omega }\to (1) \\ \text{General form of a wave equation is given by} \\ x(t)=A\sin (\omega t+\varphi ) \\ \text{Comparing the same with the given equation,} \\ x(t)=5\sin (25\pi t+0) \\ \Rightarrow \omega =25\pi  \\ \text{Substituting this in equation (1)} \\ \Rightarrow T=\dfrac{2\pi }{25\pi } \\ \Rightarrow T=0.08\text{ }\sec  \\ \text{Therefore, frequency of the signal is given by} \\ f=\dfrac{1}{T} \\ \Rightarrow f=\dfrac{1}{0.08} \\ \Rightarrow f=12.5\text{ }Hz \\ \text{When period is made } \dfrac{1}{4}^{th} \text{ of its actual magnitude,} \\ \Rightarrow T'=\dfrac{T}{4}=\dfrac{0.08}{4}=0.02\text{ sec} \\ \text{New frequency is} \\ f'=\dfrac{1}{T'} \\\Rightarrow f'=\dfrac{1}{0.02}=50\text{ }Hz \\ \therefore f'=\dfrac{625}{12.5}=\dfrac{625}{f}$

One bee, say $'A'$,  produces a sound at a period of $2\text{ sec}$ and another bee, say $'B'$,  produces a sound at a period of $4\text{ sec}$. Which bee has a higher wavelength of sound? 

Relation connecting time period, frequency and wavelength is used to determine the higher wavelength. 

 $\text{Time period of bee }'A': \\ {{T}_{A}}=2\text{ sec} \\ \text{Frequency of vibration of bee }'A': \\ {{f}_{A}}=\dfrac{1}{{{T}_{A}}} \\ \Rightarrow {{f}_{A}}=\dfrac{1}{2} \\ \Rightarrow {{f}_{A}}=0.5\text{ }Hz \\ \text{Wavelength of vibration of bee }'A': \\ {{\lambda }_{A}}=\dfrac{v}{0.5}=2V\text{ m}\to \text{(1)} \\ \text{Time period of bee }'B': \\ {{T}_{B}}=4\text{ sec} \\ \text{Frequency of vibration of bee }'B': \\ {{f}_{B}}=\dfrac{1}{{{T}_{B}}} \\ \Rightarrow {{f}_{B}}=\dfrac{1}{4} \\ \Rightarrow {{f}_{B}}=0.25\text{ }Hz \\ \text{Wavelength of vibration of bee }'A': \\ {{\lambda }_{B}}=\dfrac{v}{0.25}=4V\text{ m}\to \text{(2)} \\ \text{From (1) and (2),} \\ \therefore {{\lambda }_{B}}>{{\lambda }_{A}}$

A pendulum initially of length $10\text{ cm}$  swings through one cycle and its period is measured. Now, the length of the pendulum has been increased to $20\text{ cm}$ . Find the ratio of the frequencies before and after. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/313af918-aa2f-48ce-9f76-9a9c40a2a6eb8309539763181624695.png"/>

Relation between frequency and time period is utilized to determine the required ratio. 

 $\text{Time period of pendulum is given by} \\ T=2\pi \sqrt{\dfrac{l}{g}} \\ \text{Time period of pendulum before:} \\ \Rightarrow {{T}_{before}}=2\pi \sqrt{\dfrac{0.1}{9.8}} \\ \Rightarrow {{T}_{before}}=0.63\text{ }\sec  \\ \text{Frequency before:} \\ {{f}_{before}}=\dfrac{1}{0.63} \\ \Rightarrow {{f}_{before}}=1.58\text{ }Hz \\ \text{Time period of pendulum after:} \\ \Rightarrow {{T}_{after}}=2\pi \sqrt{\dfrac{0.2}{9.8}} \\ \Rightarrow {{T}_{after}}=0.89\text{ }\sec \\ \text{Frequency after:} \\ {{f}_{after}}=\dfrac{1}{0.89} \\ \Rightarrow {{f}_{after}}=1.11\text{ }Hz \\ \text{ratio, }\dfrac{{{f}_{before}}}{{{f}_{after}}}=\dfrac{1.58}{1.11} \\ \therefore \text{ratio}=1.42$

The angular frequency of a string of a guitar is $4\text{ rad }{{\text{s}}^{-1}}$. What happens to its frequency when this angular frequency is halved?

 Substitute the new value of angular frequency in the expression of the time period in order to arrive at final answer.

 $\text{Angular frequency = 4 rad }{{\text{s}}^{-1}} \\ \text{Time period is given by} \\ T=\dfrac{2\pi }{\omega } \\ \text{When angular frequency is halved, time period becomes} \\ T'=\dfrac{2\pi }{\left( \dfrac{\omega }{2} \right)} \\ \Rightarrow T'=\dfrac{4\pi }{\omega } \\ \Rightarrow T'=\dfrac{4\pi }{4}=\pi \text{ sec} \\ \text{Now, frequency is given by} \\ f'=\dfrac{1}{T'} \\ \therefore f'={{\pi }^{-1}}\text{ Hz}$