Question

A tuning fork of frequency 480hz produces 10 beats per second when sounded with a vibrating sonometer string what must have been the frequency of string if a slight increase in tension produces fewer beats per second than before

Answer 1

n1= 480 Hz    b= 10 b/sec      

n2 →480 +10  = 490 Hz

n2→480 -10  = 470 Hz  

  T increases Slightly = 490 Hz

Answer 2

Answer:

The frequency of string will be 490 Hz.

Explanation:

Given that,

Frequency n_{1}= 480 Hz

Number of beat b= 10 b/s

The frequency is directly proportional to the square root of tension.

If a slight increase in tension produces fewer beats per second than before

The frequency of te string will be

$n_{2}=n_{1}+b$...(I)

$n_{2}=480+10=490\ Hz$

$n_{2}=n_{1}+b$....(II)

$n_{2}=480-10=470 \Hz$

Hence, The frequency of string will be 490 Hz.