Two waves represented by y1 - 3 sin2000 - t-50x andy2 - 3sin2008 - t -50x are approaching each other- The number of beats heard per second will beAssume SI units

Given: y_1 = 3 sin(2000π t 50x ) y_2 = 3 sin(2008 π t+50x) Frequency of the first wave, f_1 = ω_12 π = 2000 π2 π = 1000 Hz and for the second wave, nbsp; n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Physics

Beat Frequency

Two waves rep...

Question

Two waves represented by $y_{1} = 3 sin (2000 π t - 50 x)$ and
$y_{2} = 3 sin (2008 π t + 50 x)$ are approaching each other. The number of beats heard per second will be
(Assume SI units)

8 Hz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 Hz

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 Hz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 Hz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $4 Hz$
Given:
$y_{1} = 3 sin (2000 π t - 50 x)$
$y_{2} = 3 sin (2008 π t + 50 x)$
Frequency of the first wave,
$f_{1} = \frac{ω_{1}}{2 π} = \frac{2000 π}{2 π} = 1000 Hz$
and for the second wave,
$f_{2} = \frac{ω_{2}}{2 π} = \frac{2008 π}{2 π} = 1004 Hz$
Now, beat frequency,
$f_{b} = | f_{1} - f_{2} | = 4 Hz$
$\Rightarrow 4$ beats per second.

Suggest Corrections

Similar questions

Q. Two waves represented by

y_{1} = 3 sin (2000 π t - 50 x)

and

y_{2} = 3 sin (2008 π t + 50 x)

are approaching each other. The number of beats heard per second will be
(Assume SI units)

Q. Beats are produced by two waves

y_{1} = a s i n 200 π t

and

y_{2} = a s i n 208 π t

. The number of beats heard per second is

Q. Beats are produced by two waves given by

y_{1} = a sin 2000 π t

and

y_{2} = a sin 2008 π t

. The number of beats heard per second is

Q. Two sound waves are given by

y_{1} = sin (2000 π t)

and

y_{2} = sin (2008 π t)

. The number of beats heard per second is:

Q. If the equations of two sound waves are

y_{1} = 5 sin 252 π t

and

y_{2} = 5 sin 281 π t

respectively. Then the number of beats heard per second will be

Join BYJU'S Learning Program

Explore more

Beat Frequency

Standard XI Physics

Join BYJU'S Learning Program

Two waves represented by y1=3sin(2000πt−50x) and y2=3sin(2008πt+50x) are approaching each other. The number of beats heard per second will be (Assume SI units)

The correct option is B 4 HzGiven: y1=3sin(2000πt−50x) y2=3sin(2008πt+50x) Frequency of the first wave, f1=ω12π=2000π2π=1000 Hz and for the second wave, f2=ω22π=2008π2π=1004 Hz Now, beat frequency, fb=|f1−f2|=4 Hz ⇒4 beats per second.

Two waves represented by $y_{1} = 3 sin (2000 π t - 50 x)$ and
$y_{2} = 3 sin (2008 π t + 50 x)$ are approaching each other. The number of beats heard per second will be
(Assume SI units)