172216
When a wave travels in a medium, the particle displacement is given by the equation $y=a$ sin $2 \pi(b t-c x)$ where $a, b$ and $c$ are constants. The maximum particle velocity will be twice the wave velocity, if
1 $\mathrm{c}=\frac{1}{\pi \mathrm{a}}$
2 $\mathrm{c}=\pi \mathrm{a}$
3 $b=a c$
4 $\mathrm{b}=\frac{1}{\mathrm{ac}}$
5 $a=b c$
Explanation:
A Given, $\quad y=a \sin 2 \pi(b t-c x)$ or $\quad y=a \sin (2 \pi b t-2 \pi c x)$ The maximum particle velocity is twice the wave velocity $\mathrm{a} \omega=2\left(\frac{\omega}{\mathrm{k}}\right)$ $\mathrm{ak}=2$ The general wave equation- $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{k}=2 \pi \mathrm{c}, \omega=2 \pi \mathrm{b}, \text { Amplitude }=\mathrm{a}$ From Eq. (i), $\therefore \mathrm{a} 2 \pi \mathrm{c} =2 (\because \mathrm{k}=2 \pi \mathrm{c})$ $\text { or } \mathrm{c} =\frac{1}{\pi \mathrm{a}}$
BCECE- 2018
WAVES
172217
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is $0.14 \mathrm{~s}$. The frequency of the wave is
1 $0.42 \mathrm{~Hz}$
2 $2.75 \mathrm{~Hz}$
3 $1.79 \mathrm{~Hz}$
4 $0.56 \mathrm{~Hz}$
5 $3.5 \mathrm{~Hz}$
Explanation:
C Given, $\frac{T}{4}=0.14 \mathrm{sec}$ Time taken by the wave to one wavelength Time period $(\mathrm{T})=0.14 \times 4=0.56 \mathrm{~s}$ Frequency- $\mathrm{f}_{\mathrm{c}}=\frac{1}{\mathrm{~T}}$ $\mathrm{f}_{\mathrm{c}}=\frac{1}{0.56}$ $\mathrm{f}=\frac{100}{56}$ $\mathrm{f}=1.79 \mathrm{~Hz}$
Kerala CEE - 2009
WAVES
172219
The speed of a wave is $360 \mathrm{~m} / \mathrm{s}$ and frequency is 500 Hz. Phase difference between two consecutive particles is $60^{\circ}$, then path difference between them will be :
1 $0.72 \mathrm{~cm}$
2 $120 \mathrm{~cm}$
3 $12 \mathrm{~cm}$
4 $7.2 \mathrm{~cm}$
Explanation:
C Given, Speed of wave $=360 \mathrm{~m} / \mathrm{s}=36000 \mathrm{~cm} / \mathrm{sec}$ $\text { Frequency }=500 \mathrm{~Hz}$ Phase difference between two consecutive particles $=$ $60^{0}$ Wavelength $=\frac{\text { velocity }}{\text { frequency }}=\frac{36000}{500}=72$ Wavelength between two consecutive particles which phase difference $=60^{\circ}$ $\frac{\lambda}{2 \pi} \times \Delta \phi=\frac{60}{360} \times \lambda$ So, distance between two particles is $\mathrm{d}=\frac{60}{360} \times 72=12 \mathrm{~cm}$
BCECE-2005
WAVES
172220
Wavelength of wave is a distance between two particles which are differing in phase by
1 $\pi$
2 $2 \pi$
3 $\frac{2 \pi}{3}$
4 $\frac{\pi}{3}$
Explanation:
B From question, Wave length of wave is a distance between two particles which are differing in phase by $2 \pi$. $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}$ $\text { If } \Delta \mathrm{x}=\lambda$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \lambda$ $\Delta \phi=2 \pi$
172216
When a wave travels in a medium, the particle displacement is given by the equation $y=a$ sin $2 \pi(b t-c x)$ where $a, b$ and $c$ are constants. The maximum particle velocity will be twice the wave velocity, if
1 $\mathrm{c}=\frac{1}{\pi \mathrm{a}}$
2 $\mathrm{c}=\pi \mathrm{a}$
3 $b=a c$
4 $\mathrm{b}=\frac{1}{\mathrm{ac}}$
5 $a=b c$
Explanation:
A Given, $\quad y=a \sin 2 \pi(b t-c x)$ or $\quad y=a \sin (2 \pi b t-2 \pi c x)$ The maximum particle velocity is twice the wave velocity $\mathrm{a} \omega=2\left(\frac{\omega}{\mathrm{k}}\right)$ $\mathrm{ak}=2$ The general wave equation- $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{k}=2 \pi \mathrm{c}, \omega=2 \pi \mathrm{b}, \text { Amplitude }=\mathrm{a}$ From Eq. (i), $\therefore \mathrm{a} 2 \pi \mathrm{c} =2 (\because \mathrm{k}=2 \pi \mathrm{c})$ $\text { or } \mathrm{c} =\frac{1}{\pi \mathrm{a}}$
BCECE- 2018
WAVES
172217
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is $0.14 \mathrm{~s}$. The frequency of the wave is
1 $0.42 \mathrm{~Hz}$
2 $2.75 \mathrm{~Hz}$
3 $1.79 \mathrm{~Hz}$
4 $0.56 \mathrm{~Hz}$
5 $3.5 \mathrm{~Hz}$
Explanation:
C Given, $\frac{T}{4}=0.14 \mathrm{sec}$ Time taken by the wave to one wavelength Time period $(\mathrm{T})=0.14 \times 4=0.56 \mathrm{~s}$ Frequency- $\mathrm{f}_{\mathrm{c}}=\frac{1}{\mathrm{~T}}$ $\mathrm{f}_{\mathrm{c}}=\frac{1}{0.56}$ $\mathrm{f}=\frac{100}{56}$ $\mathrm{f}=1.79 \mathrm{~Hz}$
Kerala CEE - 2009
WAVES
172219
The speed of a wave is $360 \mathrm{~m} / \mathrm{s}$ and frequency is 500 Hz. Phase difference between two consecutive particles is $60^{\circ}$, then path difference between them will be :
1 $0.72 \mathrm{~cm}$
2 $120 \mathrm{~cm}$
3 $12 \mathrm{~cm}$
4 $7.2 \mathrm{~cm}$
Explanation:
C Given, Speed of wave $=360 \mathrm{~m} / \mathrm{s}=36000 \mathrm{~cm} / \mathrm{sec}$ $\text { Frequency }=500 \mathrm{~Hz}$ Phase difference between two consecutive particles $=$ $60^{0}$ Wavelength $=\frac{\text { velocity }}{\text { frequency }}=\frac{36000}{500}=72$ Wavelength between two consecutive particles which phase difference $=60^{\circ}$ $\frac{\lambda}{2 \pi} \times \Delta \phi=\frac{60}{360} \times \lambda$ So, distance between two particles is $\mathrm{d}=\frac{60}{360} \times 72=12 \mathrm{~cm}$
BCECE-2005
WAVES
172220
Wavelength of wave is a distance between two particles which are differing in phase by
1 $\pi$
2 $2 \pi$
3 $\frac{2 \pi}{3}$
4 $\frac{\pi}{3}$
Explanation:
B From question, Wave length of wave is a distance between two particles which are differing in phase by $2 \pi$. $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}$ $\text { If } \Delta \mathrm{x}=\lambda$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \lambda$ $\Delta \phi=2 \pi$
172216
When a wave travels in a medium, the particle displacement is given by the equation $y=a$ sin $2 \pi(b t-c x)$ where $a, b$ and $c$ are constants. The maximum particle velocity will be twice the wave velocity, if
1 $\mathrm{c}=\frac{1}{\pi \mathrm{a}}$
2 $\mathrm{c}=\pi \mathrm{a}$
3 $b=a c$
4 $\mathrm{b}=\frac{1}{\mathrm{ac}}$
5 $a=b c$
Explanation:
A Given, $\quad y=a \sin 2 \pi(b t-c x)$ or $\quad y=a \sin (2 \pi b t-2 \pi c x)$ The maximum particle velocity is twice the wave velocity $\mathrm{a} \omega=2\left(\frac{\omega}{\mathrm{k}}\right)$ $\mathrm{ak}=2$ The general wave equation- $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{k}=2 \pi \mathrm{c}, \omega=2 \pi \mathrm{b}, \text { Amplitude }=\mathrm{a}$ From Eq. (i), $\therefore \mathrm{a} 2 \pi \mathrm{c} =2 (\because \mathrm{k}=2 \pi \mathrm{c})$ $\text { or } \mathrm{c} =\frac{1}{\pi \mathrm{a}}$
BCECE- 2018
WAVES
172217
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is $0.14 \mathrm{~s}$. The frequency of the wave is
1 $0.42 \mathrm{~Hz}$
2 $2.75 \mathrm{~Hz}$
3 $1.79 \mathrm{~Hz}$
4 $0.56 \mathrm{~Hz}$
5 $3.5 \mathrm{~Hz}$
Explanation:
C Given, $\frac{T}{4}=0.14 \mathrm{sec}$ Time taken by the wave to one wavelength Time period $(\mathrm{T})=0.14 \times 4=0.56 \mathrm{~s}$ Frequency- $\mathrm{f}_{\mathrm{c}}=\frac{1}{\mathrm{~T}}$ $\mathrm{f}_{\mathrm{c}}=\frac{1}{0.56}$ $\mathrm{f}=\frac{100}{56}$ $\mathrm{f}=1.79 \mathrm{~Hz}$
Kerala CEE - 2009
WAVES
172219
The speed of a wave is $360 \mathrm{~m} / \mathrm{s}$ and frequency is 500 Hz. Phase difference between two consecutive particles is $60^{\circ}$, then path difference between them will be :
1 $0.72 \mathrm{~cm}$
2 $120 \mathrm{~cm}$
3 $12 \mathrm{~cm}$
4 $7.2 \mathrm{~cm}$
Explanation:
C Given, Speed of wave $=360 \mathrm{~m} / \mathrm{s}=36000 \mathrm{~cm} / \mathrm{sec}$ $\text { Frequency }=500 \mathrm{~Hz}$ Phase difference between two consecutive particles $=$ $60^{0}$ Wavelength $=\frac{\text { velocity }}{\text { frequency }}=\frac{36000}{500}=72$ Wavelength between two consecutive particles which phase difference $=60^{\circ}$ $\frac{\lambda}{2 \pi} \times \Delta \phi=\frac{60}{360} \times \lambda$ So, distance between two particles is $\mathrm{d}=\frac{60}{360} \times 72=12 \mathrm{~cm}$
BCECE-2005
WAVES
172220
Wavelength of wave is a distance between two particles which are differing in phase by
1 $\pi$
2 $2 \pi$
3 $\frac{2 \pi}{3}$
4 $\frac{\pi}{3}$
Explanation:
B From question, Wave length of wave is a distance between two particles which are differing in phase by $2 \pi$. $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}$ $\text { If } \Delta \mathrm{x}=\lambda$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \lambda$ $\Delta \phi=2 \pi$
172216
When a wave travels in a medium, the particle displacement is given by the equation $y=a$ sin $2 \pi(b t-c x)$ where $a, b$ and $c$ are constants. The maximum particle velocity will be twice the wave velocity, if
1 $\mathrm{c}=\frac{1}{\pi \mathrm{a}}$
2 $\mathrm{c}=\pi \mathrm{a}$
3 $b=a c$
4 $\mathrm{b}=\frac{1}{\mathrm{ac}}$
5 $a=b c$
Explanation:
A Given, $\quad y=a \sin 2 \pi(b t-c x)$ or $\quad y=a \sin (2 \pi b t-2 \pi c x)$ The maximum particle velocity is twice the wave velocity $\mathrm{a} \omega=2\left(\frac{\omega}{\mathrm{k}}\right)$ $\mathrm{ak}=2$ The general wave equation- $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{k}=2 \pi \mathrm{c}, \omega=2 \pi \mathrm{b}, \text { Amplitude }=\mathrm{a}$ From Eq. (i), $\therefore \mathrm{a} 2 \pi \mathrm{c} =2 (\because \mathrm{k}=2 \pi \mathrm{c})$ $\text { or } \mathrm{c} =\frac{1}{\pi \mathrm{a}}$
BCECE- 2018
WAVES
172217
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is $0.14 \mathrm{~s}$. The frequency of the wave is
1 $0.42 \mathrm{~Hz}$
2 $2.75 \mathrm{~Hz}$
3 $1.79 \mathrm{~Hz}$
4 $0.56 \mathrm{~Hz}$
5 $3.5 \mathrm{~Hz}$
Explanation:
C Given, $\frac{T}{4}=0.14 \mathrm{sec}$ Time taken by the wave to one wavelength Time period $(\mathrm{T})=0.14 \times 4=0.56 \mathrm{~s}$ Frequency- $\mathrm{f}_{\mathrm{c}}=\frac{1}{\mathrm{~T}}$ $\mathrm{f}_{\mathrm{c}}=\frac{1}{0.56}$ $\mathrm{f}=\frac{100}{56}$ $\mathrm{f}=1.79 \mathrm{~Hz}$
Kerala CEE - 2009
WAVES
172219
The speed of a wave is $360 \mathrm{~m} / \mathrm{s}$ and frequency is 500 Hz. Phase difference between two consecutive particles is $60^{\circ}$, then path difference between them will be :
1 $0.72 \mathrm{~cm}$
2 $120 \mathrm{~cm}$
3 $12 \mathrm{~cm}$
4 $7.2 \mathrm{~cm}$
Explanation:
C Given, Speed of wave $=360 \mathrm{~m} / \mathrm{s}=36000 \mathrm{~cm} / \mathrm{sec}$ $\text { Frequency }=500 \mathrm{~Hz}$ Phase difference between two consecutive particles $=$ $60^{0}$ Wavelength $=\frac{\text { velocity }}{\text { frequency }}=\frac{36000}{500}=72$ Wavelength between two consecutive particles which phase difference $=60^{\circ}$ $\frac{\lambda}{2 \pi} \times \Delta \phi=\frac{60}{360} \times \lambda$ So, distance between two particles is $\mathrm{d}=\frac{60}{360} \times 72=12 \mathrm{~cm}$
BCECE-2005
WAVES
172220
Wavelength of wave is a distance between two particles which are differing in phase by
1 $\pi$
2 $2 \pi$
3 $\frac{2 \pi}{3}$
4 $\frac{\pi}{3}$
Explanation:
B From question, Wave length of wave is a distance between two particles which are differing in phase by $2 \pi$. $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}$ $\text { If } \Delta \mathrm{x}=\lambda$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \lambda$ $\Delta \phi=2 \pi$
