Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Two long strings A and B, each having linear mass density 1.2 × 10-2 kg m-1, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A?

Two long strings A and B, each having linear mass density 1.2 × 10-2 kg m-1, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A?

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
Sol. . mA = 1.2 * 10–2 kg/m, TA = 4.8 N ⇒ VA = √T /m = 20 m/s mB = 1.2 * 10–2 kg/m, TB = 7.5 N ⇒ VB = T /m = 25 m/s t = 0 in string A t1 = 0 + 20 ms = 20 * 10–3 = 0.02 sec In 0.02 sec A has travelled 20 * 0.02 = 0.4 mt Relative speed between A and B = 25 – 20 = 5 m/s Time taken for B for overtake A = s/v = 0.4/5 = 0.08 sec

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free