Two tuning forks of frequency 250Hz and 256Hz produce beats. If a maximum intensity is observed at t = 0. At what time instant(s), the minimum intensity is observed at the same place?

$\frac{1}{18}$ sec.

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

$\frac{1}{4}$ sec.

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

$\frac{1}{3}$ sec.

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

$\frac{1}{12}$ sec.

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

Open in App

Solution

The correct options are
B
$\frac{1}{4}$ sec.

D
$\frac{1}{12}$ sec.

6 beats per second. So, Time period of oscillations is $T = \frac{1}{6} s$

If maximum is at t = 0, then minima is at $\frac{T}{2} (\frac{1}{12} s), \frac{3 T}{2} (\frac{1}{4} s)$

Suggest Corrections

Similar questions

Q. Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0m s⁻¹ so that he approaches one tuning fork and recedes from the other figure. Find the beat frequency observed by the listener. Speed of sound in air = 332 m s⁻¹.

The frequencies of two sound sources are 256 Hz and 260 Hz. At t = 0, the intensity of sound is maximum. Then the phase difference at the time t = $\frac{1}{16}$ sec will be

Q. Given are two tuning forks near one another, one of them is of unknown frequency and the other is of frequency

591 Hz

. We can hear beats of maximal intensity

I_{0}, 5

times each second. Then -

Q. Two tuning forks of frequencies 250 and 256 Hz produce beats. If maximum is produced now, the time after which the minimum is produced at that same place is:

Q. Frequency of tuning fork

A

256 H z .

It produces four beats/sec with tuning fork

B .

When wax is applied at tuning fork

B

then

6

beats/sec are heard. By reducing little amount of wax

4

beats/sec are heard. Frequency of

B

is :

Join BYJU'S Learning Program

Two tuning forks of frequency 250Hz and 256Hz produce beats. If a maximum intensity is observed at t = 0. At what time instant(s), the minimum intensity is observed at the same place?

The correct options are B 14 sec. D 112 sec. 6 beats per second. So, Time period of oscillations is T=16s If maximum is at t = 0, then minima is at T2(112s),3T2(14s)

The correct options are
B
$\frac{1}{4}$ sec.

D
$\frac{1}{12}$ sec.

6 beats per second. So, Time period of oscillations is $T = \frac{1}{6} s$

If maximum is at t = 0, then minima is at $\frac{T}{2} (\frac{1}{12} s), \frac{3 T}{2} (\frac{1}{4} s)$