172261
A point moves along $x$-axis according to the law $x=\operatorname{asin}^{2}\left(\omega t-\frac{\pi}{4}\right)$. The amplitude and time period of oscillation of the point are respectively:
1 $\mathrm{a}, \frac{\pi}{2 \omega}$
2 $\frac{\mathrm{a}}{2}, \frac{\pi}{\omega}$
3 $\frac{\mathrm{a}}{2}, \frac{\pi}{2 \omega}$
4 $\mathrm{a}, \frac{\pi}{\omega}$
Explanation:
B We know that, Wave velocity $=\frac{\text { cofficient of } t}{\text { cofficient of } x}=\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=\lambda f$ Again, maximum particle velocity Given, $\mathrm{v}_{\max }=\omega \mathrm{A}=2 \pi \mathrm{fy}_{0}$ $\mathrm{v}_{\max } =4 \mathrm{v}$ $2 \pi f \mathrm{y}_{0} =4 \mathrm{f} \lambda$ $\lambda =\frac{\pi \mathrm{y}_{0}}{2}$
SCRA-2011]**#118. A transverse wave is represented by the equation $y=y_{0} \sin 2 \pi\left[f t-\left(\frac{x}{\lambda}\right)\right]$ The maximum particle velocity is equal to four times the wave velocity if
WAVES
172275
If a wave displacement is $y=0.5 \sin (0.1 \mathrm{x}+$ $0.4 t)$, where all quantities are in S.I. units, the time period is
1 $2.5 \mathrm{~s}$
2 $0.4 \mathrm{~s}$
3 $0.1 \mathrm{~s}$
4 $15.7 \mathrm{~s}$
Explanation:
D Given wave equation is - $\mathrm{y}=0.5 \sin (0.1 \mathrm{x}+0.4 \mathrm{t})$ Comparing the given equation with $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+$ $\omega t)-$ $\omega=0.4$ Now, time period $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{0.4}$ $\mathrm{T}=5 \pi=5 \times 3.14=15.7 \mathrm{~s}$
J and K CET- 1998
WAVES
172263
The equation of a wave is given by $y=10$ $\sin \left(\frac{2 \pi}{45} t+\alpha\right)$. If the displacement is $5 \mathrm{~cm}$ at $\mathrm{t}=$ 0 , then the total phase at $t=7.5 \mathrm{~s}$ is :
1 $\frac{\pi}{3}$
2 $\frac{\pi}{2}$
3 $\frac{\pi}{6}$
4 $\pi$
Explanation:
B Given, Equation of wave is, $y=10 \sin \left(\frac{2 \pi}{45} t+\alpha\right)$ At, $\mathrm{t}=0, \mathrm{y}=5 \mathrm{~cm}$ $\therefore \quad 5=10 \sin \alpha$ $\frac{1}{2}=\sin \alpha$ Or $\quad \sin \left(\frac{\pi}{6}\right)=\sin \alpha$ $\alpha=\frac{\pi}{6}$ Hence, the total phase at $\mathrm{t}=7.5 \mathrm{~s}=\left(\frac{15}{2} \mathrm{~s}\right)$ is- $\phi=\frac{2 \pi}{45} \times \frac{15}{2}+\alpha$ $\phi=\frac{\pi}{3}+\frac{\pi}{6}$ $\phi=\frac{3 \pi}{6}=\frac{\pi}{2}$
Karnataka CET-2011
WAVES
172264
$y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is :
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B Given that, $y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $y=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)$ Comparing the equation to the standard equation of a progressive wave, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}$ $\mathrm{k}=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}$ Velocity of the wave, $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{\frac{\pi}{2}}{\frac{\pi}{4}}=2 \mathrm{~ms}^{-1}$ Distance travelled by wave in time $\mathrm{t}$ is $(\mathrm{s})=\mathrm{v} \times \mathrm{t}=2 \times$ $5=10 \mathrm{~m}$ Total distance travelled $=10 \mathrm{~m}$
172261
A point moves along $x$-axis according to the law $x=\operatorname{asin}^{2}\left(\omega t-\frac{\pi}{4}\right)$. The amplitude and time period of oscillation of the point are respectively:
1 $\mathrm{a}, \frac{\pi}{2 \omega}$
2 $\frac{\mathrm{a}}{2}, \frac{\pi}{\omega}$
3 $\frac{\mathrm{a}}{2}, \frac{\pi}{2 \omega}$
4 $\mathrm{a}, \frac{\pi}{\omega}$
Explanation:
B We know that, Wave velocity $=\frac{\text { cofficient of } t}{\text { cofficient of } x}=\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=\lambda f$ Again, maximum particle velocity Given, $\mathrm{v}_{\max }=\omega \mathrm{A}=2 \pi \mathrm{fy}_{0}$ $\mathrm{v}_{\max } =4 \mathrm{v}$ $2 \pi f \mathrm{y}_{0} =4 \mathrm{f} \lambda$ $\lambda =\frac{\pi \mathrm{y}_{0}}{2}$
SCRA-2011]**#118. A transverse wave is represented by the equation $y=y_{0} \sin 2 \pi\left[f t-\left(\frac{x}{\lambda}\right)\right]$ The maximum particle velocity is equal to four times the wave velocity if
WAVES
172275
If a wave displacement is $y=0.5 \sin (0.1 \mathrm{x}+$ $0.4 t)$, where all quantities are in S.I. units, the time period is
1 $2.5 \mathrm{~s}$
2 $0.4 \mathrm{~s}$
3 $0.1 \mathrm{~s}$
4 $15.7 \mathrm{~s}$
Explanation:
D Given wave equation is - $\mathrm{y}=0.5 \sin (0.1 \mathrm{x}+0.4 \mathrm{t})$ Comparing the given equation with $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+$ $\omega t)-$ $\omega=0.4$ Now, time period $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{0.4}$ $\mathrm{T}=5 \pi=5 \times 3.14=15.7 \mathrm{~s}$
J and K CET- 1998
WAVES
172263
The equation of a wave is given by $y=10$ $\sin \left(\frac{2 \pi}{45} t+\alpha\right)$. If the displacement is $5 \mathrm{~cm}$ at $\mathrm{t}=$ 0 , then the total phase at $t=7.5 \mathrm{~s}$ is :
1 $\frac{\pi}{3}$
2 $\frac{\pi}{2}$
3 $\frac{\pi}{6}$
4 $\pi$
Explanation:
B Given, Equation of wave is, $y=10 \sin \left(\frac{2 \pi}{45} t+\alpha\right)$ At, $\mathrm{t}=0, \mathrm{y}=5 \mathrm{~cm}$ $\therefore \quad 5=10 \sin \alpha$ $\frac{1}{2}=\sin \alpha$ Or $\quad \sin \left(\frac{\pi}{6}\right)=\sin \alpha$ $\alpha=\frac{\pi}{6}$ Hence, the total phase at $\mathrm{t}=7.5 \mathrm{~s}=\left(\frac{15}{2} \mathrm{~s}\right)$ is- $\phi=\frac{2 \pi}{45} \times \frac{15}{2}+\alpha$ $\phi=\frac{\pi}{3}+\frac{\pi}{6}$ $\phi=\frac{3 \pi}{6}=\frac{\pi}{2}$
Karnataka CET-2011
WAVES
172264
$y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is :
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B Given that, $y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $y=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)$ Comparing the equation to the standard equation of a progressive wave, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}$ $\mathrm{k}=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}$ Velocity of the wave, $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{\frac{\pi}{2}}{\frac{\pi}{4}}=2 \mathrm{~ms}^{-1}$ Distance travelled by wave in time $\mathrm{t}$ is $(\mathrm{s})=\mathrm{v} \times \mathrm{t}=2 \times$ $5=10 \mathrm{~m}$ Total distance travelled $=10 \mathrm{~m}$
172261
A point moves along $x$-axis according to the law $x=\operatorname{asin}^{2}\left(\omega t-\frac{\pi}{4}\right)$. The amplitude and time period of oscillation of the point are respectively:
1 $\mathrm{a}, \frac{\pi}{2 \omega}$
2 $\frac{\mathrm{a}}{2}, \frac{\pi}{\omega}$
3 $\frac{\mathrm{a}}{2}, \frac{\pi}{2 \omega}$
4 $\mathrm{a}, \frac{\pi}{\omega}$
Explanation:
B We know that, Wave velocity $=\frac{\text { cofficient of } t}{\text { cofficient of } x}=\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=\lambda f$ Again, maximum particle velocity Given, $\mathrm{v}_{\max }=\omega \mathrm{A}=2 \pi \mathrm{fy}_{0}$ $\mathrm{v}_{\max } =4 \mathrm{v}$ $2 \pi f \mathrm{y}_{0} =4 \mathrm{f} \lambda$ $\lambda =\frac{\pi \mathrm{y}_{0}}{2}$
SCRA-2011]**#118. A transverse wave is represented by the equation $y=y_{0} \sin 2 \pi\left[f t-\left(\frac{x}{\lambda}\right)\right]$ The maximum particle velocity is equal to four times the wave velocity if
WAVES
172275
If a wave displacement is $y=0.5 \sin (0.1 \mathrm{x}+$ $0.4 t)$, where all quantities are in S.I. units, the time period is
1 $2.5 \mathrm{~s}$
2 $0.4 \mathrm{~s}$
3 $0.1 \mathrm{~s}$
4 $15.7 \mathrm{~s}$
Explanation:
D Given wave equation is - $\mathrm{y}=0.5 \sin (0.1 \mathrm{x}+0.4 \mathrm{t})$ Comparing the given equation with $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+$ $\omega t)-$ $\omega=0.4$ Now, time period $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{0.4}$ $\mathrm{T}=5 \pi=5 \times 3.14=15.7 \mathrm{~s}$
J and K CET- 1998
WAVES
172263
The equation of a wave is given by $y=10$ $\sin \left(\frac{2 \pi}{45} t+\alpha\right)$. If the displacement is $5 \mathrm{~cm}$ at $\mathrm{t}=$ 0 , then the total phase at $t=7.5 \mathrm{~s}$ is :
1 $\frac{\pi}{3}$
2 $\frac{\pi}{2}$
3 $\frac{\pi}{6}$
4 $\pi$
Explanation:
B Given, Equation of wave is, $y=10 \sin \left(\frac{2 \pi}{45} t+\alpha\right)$ At, $\mathrm{t}=0, \mathrm{y}=5 \mathrm{~cm}$ $\therefore \quad 5=10 \sin \alpha$ $\frac{1}{2}=\sin \alpha$ Or $\quad \sin \left(\frac{\pi}{6}\right)=\sin \alpha$ $\alpha=\frac{\pi}{6}$ Hence, the total phase at $\mathrm{t}=7.5 \mathrm{~s}=\left(\frac{15}{2} \mathrm{~s}\right)$ is- $\phi=\frac{2 \pi}{45} \times \frac{15}{2}+\alpha$ $\phi=\frac{\pi}{3}+\frac{\pi}{6}$ $\phi=\frac{3 \pi}{6}=\frac{\pi}{2}$
Karnataka CET-2011
WAVES
172264
$y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is :
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B Given that, $y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $y=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)$ Comparing the equation to the standard equation of a progressive wave, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}$ $\mathrm{k}=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}$ Velocity of the wave, $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{\frac{\pi}{2}}{\frac{\pi}{4}}=2 \mathrm{~ms}^{-1}$ Distance travelled by wave in time $\mathrm{t}$ is $(\mathrm{s})=\mathrm{v} \times \mathrm{t}=2 \times$ $5=10 \mathrm{~m}$ Total distance travelled $=10 \mathrm{~m}$
172261
A point moves along $x$-axis according to the law $x=\operatorname{asin}^{2}\left(\omega t-\frac{\pi}{4}\right)$. The amplitude and time period of oscillation of the point are respectively:
1 $\mathrm{a}, \frac{\pi}{2 \omega}$
2 $\frac{\mathrm{a}}{2}, \frac{\pi}{\omega}$
3 $\frac{\mathrm{a}}{2}, \frac{\pi}{2 \omega}$
4 $\mathrm{a}, \frac{\pi}{\omega}$
Explanation:
B We know that, Wave velocity $=\frac{\text { cofficient of } t}{\text { cofficient of } x}=\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=\lambda f$ Again, maximum particle velocity Given, $\mathrm{v}_{\max }=\omega \mathrm{A}=2 \pi \mathrm{fy}_{0}$ $\mathrm{v}_{\max } =4 \mathrm{v}$ $2 \pi f \mathrm{y}_{0} =4 \mathrm{f} \lambda$ $\lambda =\frac{\pi \mathrm{y}_{0}}{2}$
SCRA-2011]**#118. A transverse wave is represented by the equation $y=y_{0} \sin 2 \pi\left[f t-\left(\frac{x}{\lambda}\right)\right]$ The maximum particle velocity is equal to four times the wave velocity if
WAVES
172275
If a wave displacement is $y=0.5 \sin (0.1 \mathrm{x}+$ $0.4 t)$, where all quantities are in S.I. units, the time period is
1 $2.5 \mathrm{~s}$
2 $0.4 \mathrm{~s}$
3 $0.1 \mathrm{~s}$
4 $15.7 \mathrm{~s}$
Explanation:
D Given wave equation is - $\mathrm{y}=0.5 \sin (0.1 \mathrm{x}+0.4 \mathrm{t})$ Comparing the given equation with $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+$ $\omega t)-$ $\omega=0.4$ Now, time period $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{0.4}$ $\mathrm{T}=5 \pi=5 \times 3.14=15.7 \mathrm{~s}$
J and K CET- 1998
WAVES
172263
The equation of a wave is given by $y=10$ $\sin \left(\frac{2 \pi}{45} t+\alpha\right)$. If the displacement is $5 \mathrm{~cm}$ at $\mathrm{t}=$ 0 , then the total phase at $t=7.5 \mathrm{~s}$ is :
1 $\frac{\pi}{3}$
2 $\frac{\pi}{2}$
3 $\frac{\pi}{6}$
4 $\pi$
Explanation:
B Given, Equation of wave is, $y=10 \sin \left(\frac{2 \pi}{45} t+\alpha\right)$ At, $\mathrm{t}=0, \mathrm{y}=5 \mathrm{~cm}$ $\therefore \quad 5=10 \sin \alpha$ $\frac{1}{2}=\sin \alpha$ Or $\quad \sin \left(\frac{\pi}{6}\right)=\sin \alpha$ $\alpha=\frac{\pi}{6}$ Hence, the total phase at $\mathrm{t}=7.5 \mathrm{~s}=\left(\frac{15}{2} \mathrm{~s}\right)$ is- $\phi=\frac{2 \pi}{45} \times \frac{15}{2}+\alpha$ $\phi=\frac{\pi}{3}+\frac{\pi}{6}$ $\phi=\frac{3 \pi}{6}=\frac{\pi}{2}$
Karnataka CET-2011
WAVES
172264
$y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is :
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B Given that, $y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $y=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)$ Comparing the equation to the standard equation of a progressive wave, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}$ $\mathrm{k}=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}$ Velocity of the wave, $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{\frac{\pi}{2}}{\frac{\pi}{4}}=2 \mathrm{~ms}^{-1}$ Distance travelled by wave in time $\mathrm{t}$ is $(\mathrm{s})=\mathrm{v} \times \mathrm{t}=2 \times$ $5=10 \mathrm{~m}$ Total distance travelled $=10 \mathrm{~m}$
