When an air column at 15° C and a tunning fork are sounded together then 4 beats per second are produced, the frequency of the fork is less then that of air column. When the temperature falls to 10° C , then the beat frequency decreases by one. The frequency of the fork will be- [ v text sound . at .0° C =332 m s -1]

Question

When an air column at 15° C and a tunning fork are sounded together then 4 beats per second are produced, the frequency of the fork is less then that of air column. When the temperature falls to 10° C , then the beat frequency decreases by one. The frequency of the fork will be- [ v text sound . at .0° C =332 m s -1]  (A) 110Hz (B) 114Hz (C) 113Hz (D) 106Hz

Tardigrade Admin · Accepted Answer

(v15/2 ℓ )-n=4 (v10/2 ℓ )-n=3⇒ (v15/ v10)=(n + 4/n + 3) but vt=v0+0.6t (332 + 9/332 + 6)=(n + 4/n + 3) n=110Hz

$110 Hz$

$114 Hz$

$113 Hz$

$106 Hz$

$\frac{v _{15}}{2 ℓ} - n = 4$
$\frac{v _{10}}{2 ℓ} - n = 3 \Rightarrow \frac{v _{15}}{v _{10}} = \frac{n + 4}{n + 3}$ but $v_{t} = v_{0} + 0.6 t$
$\frac{332 + 9}{332 + 6} = \frac{n + 4}{n + 3}$
$n = 110 Hz$

110Hz

114Hz

113Hz

106Hz

2ℓv15​​−n=4 2ℓv10​​−n=3⇒v10​v15​​=n+3n+4​ but vt​=v0​+0.6t 332+6332+9​=n+3n+4​ n=110Hz

$110 Hz$

$114 Hz$

$113 Hz$

$106 Hz$

$\frac{v _{15}}{2 ℓ} - n = 4$
$\frac{v _{10}}{2 ℓ} - n = 3 \Rightarrow \frac{v _{15}}{v _{10}} = \frac{n + 4}{n + 3}$ but $v_{t} = v_{0} + 0.6 t$
$\frac{332 + 9}{332 + 6} = \frac{n + 4}{n + 3}$
$n = 110 Hz$