The following two waves are sent in opposite directions on a horizontal string to create a standing wave in a vertical plane: y 1 ( x , t ) = ( 6 . 00 mm ) sin ( 12 . 0 π x - 300 π t ) y 2 ( x , t ) = ( 6 . 00 mm ) sin ( 12 . 0 π x + 300 π t ) with x in meters and t in seconds. An antinode is located at point A . In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

Given that, two waves are traveling in opposite direction along a string as given below, y 1 = 6 mm sin ⁡ 12 π x - 300 π t y 2 = 6 mm sin ⁡ 12 π x + 300 π t Here x is in meter and t is in seconds. Both waves are having same amplitude and frequency. Resultant wave equation of superposed waves y 1 and y 2 is, y = y 1 + y 2 = 6 mm sin ⁡ 12 π x - 300 π t + 6 mm sin ⁡ 12 π x + 300 π t y = 2 6 mm sin ⁡ 12 π x cos ⁡ 300 π t y = 12 mm sin ⁡ 12 π x cos ⁡ 300 π t . . . 1 The above equation looks like, y = 2 A sin ⁡ k x cos ⁡ ω t . . . 2 By comparing equation 1 & 2 we get, Angular wave numbers, k = 12 π m - 1 Angular frequency, ω = 300 π s - 1 . Speed of the wave is, v = ω k v = 300 π 12 π ⇒ v = 25 m s - 1 The time taken to move from maximum upward displacement to maximum downward displacement is t = T 2 = 1 2 2 π ω ⇒ t = π ω t = π 300 π ⇒ t = 1 300 s = 0 .0033 s Distance covered by each wave along the string is, x = v t ⇒ x = 25 1 300 x = 1 12 m = 0 .083 m We can write directly the distance between two adjacent anti nodes is, x = λ 2 = 1 2 2 π k ⇒ x = 1 2 2 π 12 π x = 1 12 m ≅ 0 .083 m

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IMPORTANT

Earn 100

The equation of a transverse wave travelling along a very long string is $y = 3.0 s i n (0.020 π x - 4.0 π t),$ where $x$ and $y$ are expressed in centimetres, and $t$ is in seconds. Determine

(a) the amplitude

(b) the wavelength

(c) the frequency

(d) the speed

(e) the direction of propagation of the wave

(f) the maximum transverse speed of a particle in the string.

(g) What is the transverse displacement at $x = 3.5 cm$ when $t = 0.26 s ?$

Important Questions on Waves-I

MEDIUM

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 43

What is the speed of a transverse wave in a rope of length

1.75 m

and mass

60.0 g

under a tension of

500 N

MEDIUM

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 44

The function,

y (x, t) = (15.0 cm) \cos (π x - 15 π t)

, with

x

meters

and

t

seconds

, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement

y = + 6.00 cm

MEDIUM

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 45

What are

(a) the lowest frequency

(b) the second lowest frequency

for standing waves on a wire that is $10.0 m$ long, has a mass of $100 g$ , and is stretched under a tension of $275 N$ ?

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 46

A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (in the figure shown below). The waves are of two types, transverse waves travelling at $v_{t} = 50 m s^{- 1}$ and longitudinal waves traveling at $v_{l} = 150 m s^{- 1}$ . If a sudden motion sends out such waves, a scorpion can tell the distance of the beetle from the difference $∆ t$ in the arrival times of the waves at its leg nearest the beetle. What is the time difference if the distance to the beetle is $37.5 cm ?$

Question Image

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 47

Two sinusoidal waves of the same period, with amplitudes of

5.0

and

7.0 mm

, travel in the same direction along a stretched string. They produce a resultant wave with an amplitude of

10.0 mm

. The phase constant of the

5.0 mm

wave is

0

. What is the phase constant of the

7.0 mm

wave?

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 48

A sinusoidal wave is travelling on a string with speed $40 cm s^{- 1}$ . The displacement of the particles of the string at $x = 10 cm$ varies with time according to $y = (4.0 cm) s i n [5.0 - (4.0 s^{- 1}) t] .$ The linear density of the string is $4.0 g {cm}^{- 1} .$ What are
(a) the frequency
(b) the wavelength of the wave?

If the wave equation is of the form $y (x, t) = y_{m} s i n (k x \pm ω t),$ what are
(c) $y_{m}, (d) k, (e) ω,$
(f) the correct choice of sign in front of $ω$
(g) What is the tension in the string?

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 49

The following two waves are sent in opposite directions on a horizontal string to create a standing wave in a vertical plane:

\begin{array}{l} y_{1} (x, t) = (6.00 mm) \sin (12.0 π x - 300 π t) \\ y_{2} (x, t) = (6.00 mm) \sin (12.0 π x + 300 π t) \end{array}

with

x

in meters and

t

in seconds. An antinode is located at point

A

. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 16 - Waves-I>Problems>Q 50

Four waves are to be sent along the same string, in the same direction:

\begin{array}{l} y_{1} (x, t) = (5.00 mm) \sin (4 π x - 400 π t) \\ y_{2} (x, t) = (5.00 mm) \sin (4 π x - 400 π t + 0.8 π) \\ y_{3} (x, t) = (5.00 mm) \sin (4 π x - 400 π t + π) \\ y_{4} (x, t) = (5.00 mm) \sin (4 π x - 400 π t + 1.8 π) \end{array}

What is the amplitude of the resultant wave?

Important Questions on Waves-I

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