QBManager

WAVES

172147 A particle executes S.H.M. of amplitude $A$ along $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along positive $x$ axis the displacement of particle in time $t$ . $x = A \sin (ω t + δ)$ , then the value $δ$ will be:

1

\frac{π}{6}

2

\frac{π}{3}

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

A Given that,
$x = A \sin (ω t + δ)$
At, $t = 0, x = \frac{A}{2}$
Then, $\frac{A}{2} = A \sin δ$
$\sin δ = \frac{1}{2}$
$δ = 30^{\circ}$
$δ = \frac{π}{6}$

JEE Main-10.04.2023

WAVES

172148 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$
The amplitude of the motion is

1 0.5

2

\sqrt{2}

3 1

4 2

Explanation:

B $y = \sin ω t + \cos ω t$
We know that,
$y = \sin ω t + \sin (ω t + \frac{π}{2})$
$Δ ϕ = \frac{π}{2}$
$A_{R} = \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \cos (Δ ϕ)}$
$= \sqrt{1 + 1 + 2 \cos \frac{π}{2}}$
$A_{R} = \sqrt{2}$

JEE Main-10.04.2023

WAVES

172149 Which of the following wave has the largest wave speed?

1

y (x, t) = 2 \sin (2 x - 2 t)

2

y (x, t) = 3 \sin (2 x - 3 t)

3

y (x, t) = 2 \sin (3 x - 2 t)

4

y (x, t) = 3 \sin (5 x - 2 t)

Explanation:

B (a) $y (x, t) = 2 \sin (2 x - 2 t) \dots . . (i)$
Comparing equation (i) with standard equation
$y = A \sin (kx - ω t)$
$A = 2, k = 2, ω = 2 rad / \sec$
We know that,
$Wave speed (v) = \frac{ω}{k} = \frac{2}{2} = 1 m / \sec$
(b) $y (x, t) = 3 \sin (2 x - 3 t)$
Comparing equation (ii) with standard equation
$A = 3, k = 2, ω = 3 rad / \sec$ Wave speed $(v) = \frac{ω}{k} = \frac{3}{2} = 1.5 m / \sec$
(c) $y (x, t) = 2 \sin (3 x - 2 t)$
Comparing equation (iii) with standard equation
$A = 2, k = 3, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{3} m / \sec$
(d) $y (x, t) = 3 \sin (5 x - 2 t)$
Comparing equation (iv) with standard equation
$A = 3, k = 5, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{5} rad / \sec$
Hence, option ' $b$ ' wave is the maximum wave velocity.

TS EAMCET 20.07.2022

WAVES

172150 A longitudinal wave is represented by $x = 10$ $\sin 2 π (n t - \frac{x}{λ}) c m$ . The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

1

2 π

2

5 π

3

π

4

\frac{5 π}{2}

Explanation:

B The wave equation is given by
$x = 10 \sin 2 π (nt - \frac{π}{λ})$
$x = 10 \sin (2 π n t - \frac{2 π x}{λ})$
Given that
${(V_{P})}_{max} = 4 V_{wave}$
$A ω = 4 \frac{ω}{k}$
$A = \frac{4}{k}$
$10 = \frac{4}{\frac{2 π}{λ}}$
$λ = 5 π$

JEE Main-29.06.2022

WAVES

172151 A transverse wave is represented by $y = 2 \sin$ $(ω t - k x) cm$ . The value of wavelength (in $cm$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be;

1

4 π

2

2 π

3

π

4 2

Explanation:

A Transverse wave equation is given by $y = 2 \sin (ω t - kx)$
Wave velocity $= v_{wave} = \frac{ω}{k}$
Maximum velocity $= {(v_{p})}_{max} = A ω$
Given that
${(v_{P})}_{max} = (v)_{wave}$
$A ω = \frac{ω}{k}$
$A = \frac{1}{k}$
$2 = \frac{1}{2 π / λ} (∵ k = \frac{2 π}{λ})$
$λ = 4 π$

JEE Main-26.07.2022

WAVES

172147 A particle executes S.H.M. of amplitude $A$ along $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along positive $x$ axis the displacement of particle in time $t$ . $x = A \sin (ω t + δ)$ , then the value $δ$ will be:

1

\frac{π}{6}

2

\frac{π}{3}

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

A Given that,
$x = A \sin (ω t + δ)$
At, $t = 0, x = \frac{A}{2}$
Then, $\frac{A}{2} = A \sin δ$
$\sin δ = \frac{1}{2}$
$δ = 30^{\circ}$
$δ = \frac{π}{6}$

JEE Main-10.04.2023

WAVES

172148 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$
The amplitude of the motion is

1 0.5

2

\sqrt{2}

3 1

4 2

Explanation:

B $y = \sin ω t + \cos ω t$
We know that,
$y = \sin ω t + \sin (ω t + \frac{π}{2})$
$Δ ϕ = \frac{π}{2}$
$A_{R} = \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \cos (Δ ϕ)}$
$= \sqrt{1 + 1 + 2 \cos \frac{π}{2}}$
$A_{R} = \sqrt{2}$

JEE Main-10.04.2023

WAVES

172149 Which of the following wave has the largest wave speed?

1

y (x, t) = 2 \sin (2 x - 2 t)

2

y (x, t) = 3 \sin (2 x - 3 t)

3

y (x, t) = 2 \sin (3 x - 2 t)

4

y (x, t) = 3 \sin (5 x - 2 t)

Explanation:

B (a) $y (x, t) = 2 \sin (2 x - 2 t) \dots . . (i)$
Comparing equation (i) with standard equation
$y = A \sin (kx - ω t)$
$A = 2, k = 2, ω = 2 rad / \sec$
We know that,
$Wave speed (v) = \frac{ω}{k} = \frac{2}{2} = 1 m / \sec$
(b) $y (x, t) = 3 \sin (2 x - 3 t)$
Comparing equation (ii) with standard equation
$A = 3, k = 2, ω = 3 rad / \sec$ Wave speed $(v) = \frac{ω}{k} = \frac{3}{2} = 1.5 m / \sec$
(c) $y (x, t) = 2 \sin (3 x - 2 t)$
Comparing equation (iii) with standard equation
$A = 2, k = 3, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{3} m / \sec$
(d) $y (x, t) = 3 \sin (5 x - 2 t)$
Comparing equation (iv) with standard equation
$A = 3, k = 5, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{5} rad / \sec$
Hence, option ' $b$ ' wave is the maximum wave velocity.

TS EAMCET 20.07.2022

WAVES

172150 A longitudinal wave is represented by $x = 10$ $\sin 2 π (n t - \frac{x}{λ}) c m$ . The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

1

2 π

2

5 π

3

π

4

\frac{5 π}{2}

Explanation:

B The wave equation is given by
$x = 10 \sin 2 π (nt - \frac{π}{λ})$
$x = 10 \sin (2 π n t - \frac{2 π x}{λ})$
Given that
${(V_{P})}_{max} = 4 V_{wave}$
$A ω = 4 \frac{ω}{k}$
$A = \frac{4}{k}$
$10 = \frac{4}{\frac{2 π}{λ}}$
$λ = 5 π$

JEE Main-29.06.2022

WAVES

172151 A transverse wave is represented by $y = 2 \sin$ $(ω t - k x) cm$ . The value of wavelength (in $cm$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be;

1

4 π

2

2 π

3

π

4 2

Explanation:

A Transverse wave equation is given by $y = 2 \sin (ω t - kx)$
Wave velocity $= v_{wave} = \frac{ω}{k}$
Maximum velocity $= {(v_{p})}_{max} = A ω$
Given that
${(v_{P})}_{max} = (v)_{wave}$
$A ω = \frac{ω}{k}$
$A = \frac{1}{k}$
$2 = \frac{1}{2 π / λ} (∵ k = \frac{2 π}{λ})$
$λ = 4 π$

JEE Main-26.07.2022

WAVES

172147 A particle executes S.H.M. of amplitude $A$ along $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along positive $x$ axis the displacement of particle in time $t$ . $x = A \sin (ω t + δ)$ , then the value $δ$ will be:

1

\frac{π}{6}

2

\frac{π}{3}

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

A Given that,
$x = A \sin (ω t + δ)$
At, $t = 0, x = \frac{A}{2}$
Then, $\frac{A}{2} = A \sin δ$
$\sin δ = \frac{1}{2}$
$δ = 30^{\circ}$
$δ = \frac{π}{6}$

JEE Main-10.04.2023

WAVES

172148 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$
The amplitude of the motion is

1 0.5

2

\sqrt{2}

3 1

4 2

Explanation:

B $y = \sin ω t + \cos ω t$
We know that,
$y = \sin ω t + \sin (ω t + \frac{π}{2})$
$Δ ϕ = \frac{π}{2}$
$A_{R} = \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \cos (Δ ϕ)}$
$= \sqrt{1 + 1 + 2 \cos \frac{π}{2}}$
$A_{R} = \sqrt{2}$

JEE Main-10.04.2023

WAVES

172149 Which of the following wave has the largest wave speed?

1

y (x, t) = 2 \sin (2 x - 2 t)

2

y (x, t) = 3 \sin (2 x - 3 t)

3

y (x, t) = 2 \sin (3 x - 2 t)

4

y (x, t) = 3 \sin (5 x - 2 t)

Explanation:

B (a) $y (x, t) = 2 \sin (2 x - 2 t) \dots . . (i)$
Comparing equation (i) with standard equation
$y = A \sin (kx - ω t)$
$A = 2, k = 2, ω = 2 rad / \sec$
We know that,
$Wave speed (v) = \frac{ω}{k} = \frac{2}{2} = 1 m / \sec$
(b) $y (x, t) = 3 \sin (2 x - 3 t)$
Comparing equation (ii) with standard equation
$A = 3, k = 2, ω = 3 rad / \sec$ Wave speed $(v) = \frac{ω}{k} = \frac{3}{2} = 1.5 m / \sec$
(c) $y (x, t) = 2 \sin (3 x - 2 t)$
Comparing equation (iii) with standard equation
$A = 2, k = 3, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{3} m / \sec$
(d) $y (x, t) = 3 \sin (5 x - 2 t)$
Comparing equation (iv) with standard equation
$A = 3, k = 5, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{5} rad / \sec$
Hence, option ' $b$ ' wave is the maximum wave velocity.

TS EAMCET 20.07.2022

WAVES

172150 A longitudinal wave is represented by $x = 10$ $\sin 2 π (n t - \frac{x}{λ}) c m$ . The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

1

2 π

2

5 π

3

π

4

\frac{5 π}{2}

Explanation:

B The wave equation is given by
$x = 10 \sin 2 π (nt - \frac{π}{λ})$
$x = 10 \sin (2 π n t - \frac{2 π x}{λ})$
Given that
${(V_{P})}_{max} = 4 V_{wave}$
$A ω = 4 \frac{ω}{k}$
$A = \frac{4}{k}$
$10 = \frac{4}{\frac{2 π}{λ}}$
$λ = 5 π$

JEE Main-29.06.2022

WAVES

172151 A transverse wave is represented by $y = 2 \sin$ $(ω t - k x) cm$ . The value of wavelength (in $cm$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be;

1

4 π

2

2 π

3

π

4 2

Explanation:

A Transverse wave equation is given by $y = 2 \sin (ω t - kx)$
Wave velocity $= v_{wave} = \frac{ω}{k}$
Maximum velocity $= {(v_{p})}_{max} = A ω$
Given that
${(v_{P})}_{max} = (v)_{wave}$
$A ω = \frac{ω}{k}$
$A = \frac{1}{k}$
$2 = \frac{1}{2 π / λ} (∵ k = \frac{2 π}{λ})$
$λ = 4 π$

JEE Main-26.07.2022

WAVES

172147 A particle executes S.H.M. of amplitude $A$ along $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along positive $x$ axis the displacement of particle in time $t$ . $x = A \sin (ω t + δ)$ , then the value $δ$ will be:

1

\frac{π}{6}

2

\frac{π}{3}

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

A Given that,
$x = A \sin (ω t + δ)$
At, $t = 0, x = \frac{A}{2}$
Then, $\frac{A}{2} = A \sin δ$
$\sin δ = \frac{1}{2}$
$δ = 30^{\circ}$
$δ = \frac{π}{6}$

JEE Main-10.04.2023

WAVES

172148 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$
The amplitude of the motion is

1 0.5

2

\sqrt{2}

3 1

4 2

Explanation:

B $y = \sin ω t + \cos ω t$
We know that,
$y = \sin ω t + \sin (ω t + \frac{π}{2})$
$Δ ϕ = \frac{π}{2}$
$A_{R} = \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \cos (Δ ϕ)}$
$= \sqrt{1 + 1 + 2 \cos \frac{π}{2}}$
$A_{R} = \sqrt{2}$

JEE Main-10.04.2023

WAVES

172149 Which of the following wave has the largest wave speed?

1

y (x, t) = 2 \sin (2 x - 2 t)

2

y (x, t) = 3 \sin (2 x - 3 t)

3

y (x, t) = 2 \sin (3 x - 2 t)

4

y (x, t) = 3 \sin (5 x - 2 t)

Explanation:

B (a) $y (x, t) = 2 \sin (2 x - 2 t) \dots . . (i)$
Comparing equation (i) with standard equation
$y = A \sin (kx - ω t)$
$A = 2, k = 2, ω = 2 rad / \sec$
We know that,
$Wave speed (v) = \frac{ω}{k} = \frac{2}{2} = 1 m / \sec$
(b) $y (x, t) = 3 \sin (2 x - 3 t)$
Comparing equation (ii) with standard equation
$A = 3, k = 2, ω = 3 rad / \sec$ Wave speed $(v) = \frac{ω}{k} = \frac{3}{2} = 1.5 m / \sec$
(c) $y (x, t) = 2 \sin (3 x - 2 t)$
Comparing equation (iii) with standard equation
$A = 2, k = 3, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{3} m / \sec$
(d) $y (x, t) = 3 \sin (5 x - 2 t)$
Comparing equation (iv) with standard equation
$A = 3, k = 5, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{5} rad / \sec$
Hence, option ' $b$ ' wave is the maximum wave velocity.

TS EAMCET 20.07.2022

WAVES

172150 A longitudinal wave is represented by $x = 10$ $\sin 2 π (n t - \frac{x}{λ}) c m$ . The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

1

2 π

2

5 π

3

π

4

\frac{5 π}{2}

Explanation:

B The wave equation is given by
$x = 10 \sin 2 π (nt - \frac{π}{λ})$
$x = 10 \sin (2 π n t - \frac{2 π x}{λ})$
Given that
${(V_{P})}_{max} = 4 V_{wave}$
$A ω = 4 \frac{ω}{k}$
$A = \frac{4}{k}$
$10 = \frac{4}{\frac{2 π}{λ}}$
$λ = 5 π$

JEE Main-29.06.2022

WAVES

172151 A transverse wave is represented by $y = 2 \sin$ $(ω t - k x) cm$ . The value of wavelength (in $cm$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be;

1

4 π

2

2 π

3

π

4 2

Explanation:

A Transverse wave equation is given by $y = 2 \sin (ω t - kx)$
Wave velocity $= v_{wave} = \frac{ω}{k}$
Maximum velocity $= {(v_{p})}_{max} = A ω$
Given that
${(v_{P})}_{max} = (v)_{wave}$
$A ω = \frac{ω}{k}$
$A = \frac{1}{k}$
$2 = \frac{1}{2 π / λ} (∵ k = \frac{2 π}{λ})$
$λ = 4 π$

JEE Main-26.07.2022

WAVES

172147 A particle executes S.H.M. of amplitude $A$ along $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along positive $x$ axis the displacement of particle in time $t$ . $x = A \sin (ω t + δ)$ , then the value $δ$ will be:

1

\frac{π}{6}

2

\frac{π}{3}

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

A Given that,
$x = A \sin (ω t + δ)$
At, $t = 0, x = \frac{A}{2}$
Then, $\frac{A}{2} = A \sin δ$
$\sin δ = \frac{1}{2}$
$δ = 30^{\circ}$
$δ = \frac{π}{6}$

JEE Main-10.04.2023

WAVES

172148 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$
The amplitude of the motion is

1 0.5

2

\sqrt{2}

3 1

4 2

Explanation:

B $y = \sin ω t + \cos ω t$
We know that,
$y = \sin ω t + \sin (ω t + \frac{π}{2})$
$Δ ϕ = \frac{π}{2}$
$A_{R} = \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \cos (Δ ϕ)}$
$= \sqrt{1 + 1 + 2 \cos \frac{π}{2}}$
$A_{R} = \sqrt{2}$

JEE Main-10.04.2023

WAVES

172149 Which of the following wave has the largest wave speed?

1

y (x, t) = 2 \sin (2 x - 2 t)

2

y (x, t) = 3 \sin (2 x - 3 t)

3

y (x, t) = 2 \sin (3 x - 2 t)

4

y (x, t) = 3 \sin (5 x - 2 t)

Explanation:

B (a) $y (x, t) = 2 \sin (2 x - 2 t) \dots . . (i)$
Comparing equation (i) with standard equation
$y = A \sin (kx - ω t)$
$A = 2, k = 2, ω = 2 rad / \sec$
We know that,
$Wave speed (v) = \frac{ω}{k} = \frac{2}{2} = 1 m / \sec$
(b) $y (x, t) = 3 \sin (2 x - 3 t)$
Comparing equation (ii) with standard equation
$A = 3, k = 2, ω = 3 rad / \sec$ Wave speed $(v) = \frac{ω}{k} = \frac{3}{2} = 1.5 m / \sec$
(c) $y (x, t) = 2 \sin (3 x - 2 t)$
Comparing equation (iii) with standard equation
$A = 2, k = 3, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{3} m / \sec$
(d) $y (x, t) = 3 \sin (5 x - 2 t)$
Comparing equation (iv) with standard equation
$A = 3, k = 5, ω = 2 rad / \sec$
Wave speed $(v) = \frac{ω}{k} = \frac{2}{5} rad / \sec$
Hence, option ' $b$ ' wave is the maximum wave velocity.

TS EAMCET 20.07.2022

WAVES

172150 A longitudinal wave is represented by $x = 10$ $\sin 2 π (n t - \frac{x}{λ}) c m$ . The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

1

2 π

2

5 π

3

π

4

\frac{5 π}{2}

Explanation:

B The wave equation is given by
$x = 10 \sin 2 π (nt - \frac{π}{λ})$
$x = 10 \sin (2 π n t - \frac{2 π x}{λ})$
Given that
${(V_{P})}_{max} = 4 V_{wave}$
$A ω = 4 \frac{ω}{k}$
$A = \frac{4}{k}$
$10 = \frac{4}{\frac{2 π}{λ}}$
$λ = 5 π$

JEE Main-29.06.2022

WAVES

172151 A transverse wave is represented by $y = 2 \sin$ $(ω t - k x) cm$ . The value of wavelength (in $cm$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be;

1

4 π

2

2 π

3

π

4 2

Explanation:

A Transverse wave equation is given by $y = 2 \sin (ω t - kx)$
Wave velocity $= v_{wave} = \frac{ω}{k}$
Maximum velocity $= {(v_{p})}_{max} = A ω$
Given that
${(v_{P})}_{max} = (v)_{wave}$
$A ω = \frac{ω}{k}$
$A = \frac{1}{k}$
$2 = \frac{1}{2 π / λ} (∵ k = \frac{2 π}{λ})$
$λ = 4 π$

JEE Main-26.07.2022

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