Two identical strings, A and B, have nearly the same tension. When they both vibrate in their fundamental resonant modes, there is a beat of 2 Hz. When string A is tightened slightly, to increase the tension, the beat frequency becomes 5 Hz. This means
A)that before and after tightening B has a higher frequencythan that of A
B)That before and after tightening A has a higher frequencythan B.
C)that before tightening B had a higher frequency than A, butafter tightening, A has a higher frequency than B.
d) that before tightening A had a higher frequecy than B, butafter tightening. B has a higher frequency than A.
E) none of the above.

B)That before and after tightening A has a higher frequency than B.

Explanation:

Frequency is directly proportional to the Tension on the string.

Beat frequency is the difference in frequency between two waves.

The initial beat frequency is 2Hz.

If the beat frequency increased to 5Hz (increased by 3Hz), after increasing the tension on string A, it simply means that the frequency of string A increased by 3Hz, since frequency is directly proportional to tension on string.

Therefore, if the difference in frequency between string A and string B is 2Hz, and increase in the tension on A caused the difference (beat frequency) to increase to 5Hz. It can be concluded, that before and after tightening A has a higher frequency than B

Option b is the correct answer.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length