2) A string under a tension of 129.6 N produces
10 beats/sec when it is vibrated along with a
tuning fork. When the tension is the string is
increased to 160 N it sounds in unison with same
tuning fork. Calculate fundamental freq. of tuning
fork.
(1) 100 Hz
(2) 50 Hz
(3) 150 Hz
(4) 200 Hz
Answers
The Frequency of the tuning fork is 100 Hz.
Explanation:
Given data:
T 1 =126.6 N
T2 = 160 N
2 =160 N
and b = 10 beats/sec
As, in first medium,
f1 = 1 / 2l √T1 /m= f0 - b .... (i)
In the second condition,
f2 =1 / 2l √T2 /m = f0 ..... (ii)f
On dividing Eq. (i) by Eq. (ii), we get
√T1/T2 = f0 - b/ f0
√129.6 / 160 = f0 - 10/ f0
f0 = 100 Hz
Thus the Frequency of the tuning fork is 100 Hz.
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Tuning fork produces 8 beats with length 10 and 12 cm find frequency ?
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Let f,f 1,f 2 are the fork frequency,initial and final frequencies of the string respectively.
Given : ∣f−f 1∣=10 & f=f2
We know that f ∝T
for same no.of loops
Clearly f1 is less than f2 because frequency increases on increasing tension in the wire.
We have f2 f1 = T 2T 1
f. f−10 = 160 .129.6
Solving we get f=100Hz
The fundamental frequency of tuning fork is 100Hz.