1great solution guys........cheers
A heavy string istied atone end to amoveable support and to a light string at the other end as shown. he string goes oer a puley(fixed) & supports a weight to prouce a tenion. The lowest frequency with which the heavy strin(in red coloue) resonates is 120 Hz. If the moveable support is pushed to the right by 10 cm so that the joint is placed on the pulley , what will be the minimum frequecy at which the haevy strig can resonate?
(H C V Part I page- 327 last Q)
ans(240 Hz)
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3 Answers
24Initially since end is free,so antinode formed
if support pushed to rite,a node will be formed at the place where joint formed
=> wavelength reduces by half and frequency 2 times.......(form simple 2 eqn urself if u feel the need)
.=>fnew=2finitial=240 Hz[1]
6The junction between a heavy string and a light string always acts as a free end for the heavy string hence the minimum frequency over here will be
when the block shifts by ten cms then the junction will act as a closed end or fixed end and hence the fundamental frequency is now given as
cheers!!!!!!!!!!