如图所示,实线为一列简谐波在t=0时刻的波形,虚线表示经过△t=0.2s后它的波形图象,已知T<△t<2T(T表示周期),则这列波传播速度的可能值v=______;这列波的频率可能值f=______.
由图线可直接读出波长λ=0.04m.
因为T<△t<2T
所以当波向右传播时,0.2s=
T,周期T=5 4
s,则f=4 25
=1 T
=6.25Hz,v=25 4
=λ T
=0.25m/s0.04 4 25
当波向左传播时,0.2s=
T,周期T=7 4
s,则f=4 35
=1 T
=8.75Hz,v=35 4
=λ T
=0.35m/s0.04 4 35
故答案为:0.25m/s,0.35m/s;6.25Hz,8.75Hz