A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string

A Tuning Fork Of Frequency 512 Hz Makes 4 Beats Physics Question

A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options

(a) 510 Hz
(b) 514 Hz
(c) 516 Hz
(d) 508 Hz

Correct Answer:

508 Hz

Explanation:

The frequency of the piano string = 512+-4 = 516 or 508. When the tension is increased, beat frequency decreases to 2, it means that frequency of othe string is 508 as frequency of string increases with tension.

Related Questions:

  1. The Kα X-ray of molybdenum has a wavelength of 71×10⁻¹² m. If the energy of a molybdenum
  2. The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in kg) moving in one
  3. Figure below shows two paths that may be taken by a gas to go from a state A
  4. Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
  5. If in a p-n junction, a square input signal of 10 V is applied as shown, then the output

Topics: Waves (80)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*