Question

In Figure (a), string 1 has a linear density of 4.18 g/m, and string 2 has a linear density of 8.95 g/m. They are under tension due to the hanging block of mass M = 543 g. Calculate the wave speed on (a) string 1 and (b) string 2. (Hint: When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with M2+ M2 - M) and the apparatus is rearranged as shown in Figure (b). Find (c) M1 and (d) M2 such that the wave speeds in the two strings are equal. -Suring 1 String 2 M -String 1 String 2 (a) Number i Units (b) Number Units (c) Number Units (d) Number Units

Answer 1

In fig o वर the sking 1 weight f is the mars. see that We Can ळवग्ग्क tenision striog 2. tiक two h৭IE is 2 block Ma 2 1 wave speed oo striog Ca) Now गप given by is Mg 2 LI V, = 25.23 M 0.543 x 9.8 2 x 418 x1o-3 waue speed on )Similasty Ma O.543 x g 8 2 x 8-9S xi03 17. 24 tension in string t this case ( Now T, = M,4 tension in string 2 M2 is

.. Wave speed on string 1 cking 2 respectively are Now fox wave speed be on equal e M22 M2 MI Also C Given) M M 2 putting in eq M-M ,M- M,) M, 2 M O.543 M 4t8-95x1 o. 1729 ka M, 172.9 9 PR from M2 o. 370 o O.542-O-I729 M20.370) kg 270. 1 32