Wave Motion – Quantum Study

In Resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air column …

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Q: In Resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air column in the tube is 19 cm. If the 400 Hz. tuning fork is replaced by 1600 Hz tuning fork then to get resonance, the water level in the tube should be further lowered by (take end correction = 1 cm)

(A) 5 cm

(B) 10 cm

(D) 20 cm

Click to See Solution :

Ans: (A),(C)

Sol: For first resonance with 400 Hz tuning fork

$\displaystyle L_{eq} = \frac{v}{4 f_0 }$

$\displaystyle L_{eq} = \frac{v}{4 (400) } = (19+1) = 20 cm$

Numerical

$\displaystyle \frac{v}{4 f_0 } = \frac{v}{4 (1600) } = \frac{20}{4} = 5 cm$

For Resonance

$\displaystyle L_{eq} = \frac{v}{4 f_0 } , \frac{3 v}{4 f_0 } , \frac{5 v}{4 f_0 }, \frac{7 v}{4 f_0 }….$

1 cm + L = 5 cm , 15 cm , 25 cm , 35 cm , 45 cm …..

L = 4 cm , 14 cm , 24 cm , 34 cm , 44 cm …..

water level should be further lowered by
24 – 19 = 5 cm

⇒ 34 – 19 = 15 cm

The source (S) of sound is moving constant velocity v0 as shown in diagram. An observer O listens to the sound ….

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Q: The source (S) of sound is moving constant velocity v₀ as shown in diagram. An observer O listens to the sound emitted by the source. The observed frequency of the sound

Numerical

(A) continuously decreases

(B) continuously increases

(D) first increases then decreases

Click to See Solution :

Ans: (A)

Sol: From figure , the velocity of approach (Vcosθ) decrease as the source comes closer (as θ increases).And the velocity of separation also increases as φ will decrease. Hence the frequency of sound as heared by the observed decreases continuously .

Numerical

A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode….

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Q: A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the
centre of the string is 4 mm. Distance between the two points having amplitude 2 mm is:

(A) 1 m

(B) 75 cm

(D) 50 cm

Click to See Solution :

Ans: (A)

Sol: λ = 2 l = 3 m

Equation of standing wave

y = 2A sin kx cos ωty = A as amplitude is 2A

A = 2A sin kx

$sin k x = \frac{1}{2}$

$k x = \frac{\pi}{6}$

$\frac{2\pi}{\lambda} x = \frac{\pi}{6}$

$\frac{2\pi}{3} x = \frac{\pi}{6}$

x₁ = 1/4 m

And , $k x = \frac{5 \pi}{6}$

$\frac{2\pi}{\lambda} x = \frac{5 \pi}{6}$

x₂ = 1.25 m

x₂ – x₁ = 1 m

A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis.A particle of string …

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Q: COMPREHENSION:
A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string
at x = 2 m is found at its mean position and it is moving in positive y direction at t = 1 sec. If the amplitude
of the wave, the wavelength and the angular frequency of the wave are 0.1 meter, π/4 meter and 4π rad/sec
respectively.

1 .The equation of the wave is

(A) y = 0.1 sin (4πt – 1)+ 8(x – 2))

(B) y = 0.1 sin (t-1)- (x – 2))

(D) none of these

2. The speed of particle at x = 2 m and t = 1 sec is

(A) 0.2π m/s

(B) 0.6π m/s

(D) 0

3. The instantaneous power transfer through x = 2 m and t = 1.125 sec, is
(A) 10 J/s

(B) 4π/3 J/s

(D) zero

Click to See Solution :

Ans: 1.(A) ; 2. (C) ; 3.(D)

Sol: 1 . The equation of wave moving in negative x-direction, assuming origin of position at x = 2 and origin of time
(i.e. initial time) at t = 1 sec. y = 0.1 sin (4πt + 8x)
Shifting the origin of position to left by 2m, that is, to
x = 0. Also shifting the origin of time backwards by 1 sec,
that is to t = 0 sec. y = 0.1 sin [(4πt + 8(x – 2)]

2. As given the particle at x = 2 is at mean position at t = 1 sec. its velocity v = ωA = 4π × 0.1 = 0.4 π m/s.

3. Time period of oscillation T = 2π/ω sec. Hence at t = 1.125 sec, that is, at T/4 seconds after t = 1 second, the particle is at rest at extreme position. Hence instantaneous power at x = 2 at t = 1.125 sec is zero.